\displaystyle \textbf{Question 1: }~\text{Which of the following distributions of probabilities of a random} \\ \text{variable }X\text{ arethe probability distributions?}
\displaystyle \text{(i)}\ \begin{array}{c|ccccc} X & 3 & 2 & 1 & 0 & -1\\ \hline P(X) & 0.3 & 0.2 & 0.4 & 0.1 & 0.05 \end{array}
\displaystyle \text{(ii)}\ \begin{array}{c|ccc} X & 0 & 1 & 2\\ \hline P(X) & 0.6 & 0.4 & 0.2 \end{array}
\displaystyle \text{(iii)}\ \begin{array}{c|ccccc} X & 0 & 1 & 2 & 3 & 4\\ \hline P(X) & 0.1 & 0.5 & 0.2 & 0.1 & 0.1 \end{array}
\displaystyle \text{(iv)}\ \begin{array}{c|cccc} X & 0 & 1 & 2 & 3\\ \hline P(X) & 0.3 & 0.2 & 0.4 & 0.1 \end{array}
\displaystyle \text{Answer:}
\displaystyle \text{(i)}\ P(X=3)+P(X=2)+P(X=1)+P(X=0)+P(X=-1)
\displaystyle =0.3+0.2+0.4+0.1+0.05
\displaystyle =1.05>1
\displaystyle \text{It\ is\ not\ a\ probability\ distribution\ of\ random\ variable\ }X.
\displaystyle \text{(ii)}\ P(X=0)+P(X=1)+P(X=2)
\displaystyle =0.6+0.4+0.2
\displaystyle =1.2>1
\displaystyle \text{It\ is\ not\ a\ probability\ distribution\ of\ random\ variable\ }X.
\displaystyle \text{(iii)}\ P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)
\displaystyle =0.1+0.5+0.2+0.1+0.1
\displaystyle =1
\displaystyle \text{It\ is\ a\ probability\ distribution\ of\ random\ variable\ }X.
\displaystyle \text{(iv)}\ P(X=0)+P(X=1)+P(X=2)+P(X=3)
\displaystyle =0.3+0.2+0.4+0.1
\displaystyle =1
\displaystyle \text{It\ is\ a\ probability\ distribution\ of\ random\ variable\ }X.

\displaystyle \textbf{Question 2: }~\text{A random variable }X\text{ has the following probability distribution:}
\displaystyle \begin{array}{c|cccccc} X & -2 & -1 & 0 & 1 & 2 & 3\\ \hline P(X) & 0.1 & k & 0.2 & 2k & 0.3 & k \end{array}
\displaystyle \text{Find the value of }k.
\displaystyle \text{Answer:}
\displaystyle \text{We\ know\ that\ the\ sum\ of\ probabilities\ in\ a\ probability\ distribution\ is\ always\ }1.
\displaystyle \therefore\ P(X=-2)+P(X=-1)+P(X=0)+P(X=1)+P(X=2)+P(X=3)=1
\displaystyle \Rightarrow 0.1+k+0.2+2k+0.3+k=1
\displaystyle \Rightarrow 4k+0.6=1
\displaystyle \Rightarrow 4k=0.4
\displaystyle \Rightarrow k=\frac{0.4}{4}=0.1

\displaystyle \textbf{Question 3: }~\text{A random variable }X\text{ has the following probability distribution:}
\displaystyle \begin{array}{c|ccccccccc} X & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8\\ \hline P(X) & a & 3a & 5a & 7a & 9a & 11a & 13a & 15a & 17a \end{array}
\displaystyle \text{Determine: (i) the value of }a,\ (ii)\ P(X<3),\ P(X\ge 3),\ P(0<X<5).
\displaystyle \text{Answer:}
\displaystyle \text{(i)\ Since\ the\ sum\ of\ probabilities\ in\ a\ probability\ distribution\ is\ always\ }1.
\displaystyle \therefore\ P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)+P(X=6)+P(X=7)+P(X=8)=1
\displaystyle \Rightarrow a+3a+5a+7a+9a+11a+13a+15a+17a=1
\displaystyle \Rightarrow 81a=1
\displaystyle \Rightarrow a=\frac{1}{81}
\displaystyle \text{(ii)}\ P(X<3)
\displaystyle =P(X=0)+P(X=1)+P(X=2)
\displaystyle =\frac{1}{81}+\frac{3}{81}+\frac{5}{81}
\displaystyle =\frac{9}{81}
\displaystyle =\frac{1}{9}
\displaystyle P(X\ge 3)
\displaystyle =P(X=3)+P(X=4)+P(X=5)+P(X=6)+P(X=7)+P(X=8)
\displaystyle =\frac{7}{81}+\frac{9}{81}+\frac{11}{81}+\frac{13}{81}+\frac{15}{81}+\frac{17}{81}
\displaystyle =\frac{72}{81}
\displaystyle =\frac{8}{9}
\displaystyle P(0<X<5)
\displaystyle =P(X=1)+P(X=2)+P(X=3)+P(X=4)
\displaystyle =\frac{3}{81}+\frac{5}{81}+\frac{7}{81}+\frac{9}{81}
\displaystyle =\frac{24}{81}
\displaystyle =\frac{8}{27}

\displaystyle \textbf{Question 4: }~\text{The probability distribution function of a random variable }X\text{ is given by:}
\displaystyle \begin{array}{c|ccc} x_i & 0 & 1 & 2\\ \hline p_i & 3c^3 & 4c-10c^2 & 5c-1 \end{array}
\displaystyle \text{where }c>0.\ \text{Find: (i) }c,\ (ii)\ P(X<2),\ (iii)\ P(1<X\le 2).
\displaystyle \text{Answer:}
\displaystyle \text{(i)  }
\displaystyle \text{We\ know\ that\ the\ sum\ of\ probabilities\ in\ a\ probability\ distribution\ is\ always\ }1.
\displaystyle \therefore\ P(X=0)+P(X=1)+P(X=2)=1
\displaystyle \Rightarrow 3c^{3}+4c-10c^{2}+5c-1=1
\displaystyle \Rightarrow 3c^{3}-10c^{2}+9c-2=0
\displaystyle \Rightarrow (c-1)(3c^{2}-7c+2)=0
\displaystyle \Rightarrow (c-1)(3c-1)(c-2)=0
\displaystyle \Rightarrow c=\frac{1}{3},1,2
\displaystyle \text{Neglecting\ }1\ \text{and\ }2\ \text{as\ individual\ probability\ should\ not\ be\ greater\ than\ }1.
\displaystyle \text{(ii)  }
\displaystyle P(X<2)
\displaystyle =P(X=0)+P(X=1)
\displaystyle =3c^{3}+4c-10c^{2}
\displaystyle =\frac{1}{9}+\frac{4}{3}-\frac{10}{9}
\displaystyle =\frac{1+12-10}{9}
\displaystyle =\frac{3}{9}
\displaystyle =\frac{1}{3}
\displaystyle \text{(iii)  }
\displaystyle P(1<X\le 2)
\displaystyle =P(X=2)
\displaystyle =5c-1
\displaystyle =\frac{5}{3}-1
\displaystyle =\frac{5-3}{3}
\displaystyle =\frac{2}{3}

\displaystyle \textbf{Question 5: }~\text{Let }X\text{ be a random variable which assumes values }x_1,x_2,x_3,x_4 \\ \text{such that }2P(X=x_1)=3P(X=x_2)=P(X=x_3)=5P(X=x_4).\ \text{Find} \\ \text{the probability distribution of }X.
\displaystyle \text{Answer:}
\displaystyle \text{Let}\ P(X=x_{3})=k.
\displaystyle P(X=x_{1})=\frac{k}{2}
\displaystyle P(X=x_{2})=\frac{k}{3}
\displaystyle P(X=x_{4})=\frac{k}{5}
\displaystyle \text{We\ know\ that\ the\ sum\ of\ probabilities\ in\ a\ probability\ distribution\ is\ always\ }1.
\displaystyle \therefore\ P(X=x_{1})+P(X=x_{2})+P(X=x_{3})+P(X=x_{4})=1
\displaystyle \Rightarrow \frac{k}{2}+\frac{k}{3}+k+\frac{k}{5}=1
\displaystyle \Rightarrow \frac{15k+10k+30k+6k}{30}=1
\displaystyle \Rightarrow \frac{61k}{30}=1
\displaystyle \Rightarrow k=\frac{30}{61}
\displaystyle \text{Now,}
\displaystyle x_{1}\qquad p_{1}=\frac{k}{2}=\frac{15}{61}
\displaystyle x_{2}\qquad p_{2}=\frac{k}{3}=\frac{10}{61}
\displaystyle x_{3}\qquad p_{3}=k=\frac{30}{61}
\displaystyle x_{4}\qquad p_{4}=\frac{k}{5}=\frac{6}{61}

\displaystyle \textbf{Question 6: }~\text{A random variable }X\text{ takes the values }0,1,2,3\text{ such that:}
\displaystyle P(X=0)=P(X>0)=P(X<0);\ P(X=3)=P(X=-2)=P(X=-1);\ P(X=1)=P(X=2)=P(X=3).\ \text{Obtain the probability distribution of }X.
\displaystyle \text{Answer:}
\displaystyle \text{Let}\ P(X=0)=k.
\displaystyle P(X=0)=P(X>0)=P(X<0)
\displaystyle \Rightarrow P(X>0)=k
\displaystyle P(X<0)=k
\displaystyle \therefore\ P(X=0)+P(X>0)+P(X<0)=1
\displaystyle \Rightarrow k+k+k=1
\displaystyle \Rightarrow k=\frac{1}{3}
\displaystyle \text{Now,}
\displaystyle P(X<0)=k
\displaystyle \Rightarrow P(X=-1)+P(X=-2)+P(X=-3)=k
\displaystyle \Rightarrow 3P(X=-1)=k\ \text{since}\ P(X=-1)=P(X=-2)=P(X=-3)
\displaystyle \Rightarrow P(X=-1)=\frac{k}{3}
\displaystyle \Rightarrow P(X=-1)=\frac{1}{3}\times\frac{1}{3}=\frac{1}{9}
\displaystyle \therefore\ P(X=-1)=P(X=-2)=P(X=-3)=\frac{1}{9}
\displaystyle \text{Similarly,}
\displaystyle P(X>0)=k
\displaystyle \Rightarrow P(X=1)=P(X=2)=P(X=3)=\frac{1}{9}
\displaystyle \text{Thus,\ the\ probability\ distribution\ is\ given\ by}
\displaystyle \begin{array}{c|cccccc} X_i & -3 & -2 & -1 & 1 & 2 & 3 \\ \hline P_i & \frac{1}{9} & \frac{1}{9} & \frac{1}{9} & \frac{1}{9} & \frac{1}{9} & \frac{1}{9} \end{array}

\displaystyle \textbf{Question 7: }~\text{Two cards are drawn from a well shuffled pack of 52 cards. Find the} \\ \text{probability distribution of the number of aces.}
\displaystyle \text{Answer:}
\displaystyle \text{Let } X \text{ denote the number of aces in a sample of } 2 \text{ cards drawn from a well-shuffled pack of } \\ 52 \text{ playing cards. Then } X \text{ can take the values } 0, 1 \text{ and } 2.
\displaystyle \text{Now,}
\displaystyle P(X=0)
\displaystyle =P(\text{no\ ace})
\displaystyle =\frac{{}^{48}C_{2}}{{}^{52}C_{2}}
\displaystyle =\frac{2256}{1326}
\displaystyle =\frac{188}{221}
\displaystyle P(X=1)
\displaystyle =P(\text{1\ ace})
\displaystyle =\frac{{}^{4}C_{1}\times{}^{48}C_{1}}{{}^{52}C_{2}}
\displaystyle =\frac{192}{1326}
\displaystyle =\frac{32}{221}
\displaystyle P(X=2)
\displaystyle =P(\text{2\ aces})
\displaystyle =\frac{{}^{4}C_{2}}{{}^{52}C_{2}}
\displaystyle =\frac{6}{1326}
\displaystyle =\frac{1}{221}
\displaystyle \text{Thus,\ the\ probability\ distribution\ of\ }X\ \text{is\ given\ by}
\displaystyle \begin{array}{c|ccc} X & 0 & 1 & 2 \\ \hline P(X) & \frac{188}{221} & \frac{32}{221} & \frac{1}{221} \end{array}

\displaystyle \textbf{Question 8: }~\text{Find the probability distribution of the number of heads, when three} \\ \text{coins are tossed.}
\displaystyle \text{Answer:}
\displaystyle \text{Let } X \text{ denote the number of heads in three tosses of a coin. Then } X \text{ can take the values } \\ 0, 1, 2 \text{ and } 3.
\displaystyle \text{Now,}
\displaystyle P(X=0)=P(\text{TTT})=\frac{1}{8}
\displaystyle P(X=1)=P(\text{HTT\ or\ THT\ or\ TTH})=\frac{3}{8}
\displaystyle P(X=2)=P(\text{HTH\ or\ THH\ or\ HHT})=\frac{3}{8}
\displaystyle P(X=3)=P(\text{HHH})=\frac{1}{8}
\displaystyle \text{Thus,\ the\ probability\ distribution\ of\ }X\ \text{is\ given\ by}
\displaystyle \begin{array}{c|cccc} X & 0 & 1 & 2 & 3 \\ \hline P(X) & \frac{1}{8} & \frac{3}{8} & \frac{3}{8} & \frac{1}{8} \end{array}

\displaystyle \textbf{Question 9: }~\text{Four cards are drawn simultaneously from a well shuffled pack of 52} \\ \text{playing cards. Find the probability distribution of the number of aces.}
\displaystyle \text{Answer:}
\displaystyle \text{Let } X \text{ denote the number of aces in a sample of } 4 \text{ cards drawn from a well-shuffled pack of } \\ 52 \text{ playing cards. Then } X \text{ can take the values } 0, 1, 2, 3 \text{ and } 4.
\displaystyle \text{Now,}
\displaystyle P(X=0)=P(\text{no\ ace})=\frac{{}^{48}C_{4}}{{}^{52}C_{4}}
\displaystyle P(X=1)=P(\text{1\ ace})=\frac{{}^{4}C_{1}\times{}^{48}C_{3}}{{}^{52}C_{4}}
\displaystyle P(X=2)=P(\text{2\ aces})=\frac{{}^{4}C_{2}\times{}^{48}C_{2}}{{}^{52}C_{4}}
\displaystyle P(X=3)=P(\text{3\ aces})=\frac{{}^{4}C_{3}\times{}^{48}C_{1}}{{}^{52}C_{4}}
\displaystyle P(X=4)=P(\text{4\ aces})=\frac{{}^{4}C_{4}}{{}^{52}C_{4}}
\displaystyle \text{Thus,\ the\ probability\ distribution\ of\ }X\ \text{is\ given\ by}
\displaystyle \begin{array}{c|ccccc} X & 0 & 1 & 2 & 3 & 4 \\ \hline P(X) & \frac{{}^{48}C_{4}}{{}^{52}C_{4}} & \frac{{}^{4}C_{1}\times{}^{48}C_{3}}{{}^{52}C_{4}} & \frac{{}^{4}C_{2}\times{}^{48}C_{2}}{{}^{52}C_{4}} & \frac{{}^{4}C_{3}\times{}^{48}C_{1}}{{}^{52}C_{4}} & \frac{{}^{4}C_{4}}{{}^{52}C_{4}} \end{array}

\displaystyle \textbf{Question 10: }~\text{A bag contains 4 red and 6 black balls. Three balls are drawn at random.} \\ \text{Find the probability distribution of the number of red balls.}
\displaystyle \text{Answer:}
\displaystyle \text{Let } X \text{ denote the number of red balls in a sample of } 3 \text{ balls drawn from a bag containing } \\ 4 \text{ red and } 6 \text{ black balls. Then } X \text{ can take the values } 0, 1, 2 \text{ and } 3.
\displaystyle \text{Now,}
\displaystyle P(X=0)
\displaystyle =P(\text{no\ red\ ball})
\displaystyle =\frac{{}^{6}C_{3}}{{}^{10}C_{3}}
\displaystyle =\frac{20}{120}
\displaystyle =\frac{1}{6}
\displaystyle P(X=1)
\displaystyle =P(\text{1\ red\ ball})
\displaystyle =\frac{{}^{4}C_{1}\times{}^{6}C_{2}}{{}^{10}C_{3}}
\displaystyle =\frac{60}{120}
\displaystyle =\frac{1}{2}
\displaystyle P(X=2)
\displaystyle =P(\text{2\ red\ balls})
\displaystyle =\frac{{}^{4}C_{2}\times{}^{6}C_{1}}{{}^{10}C_{3}}
\displaystyle =\frac{36}{120}
\displaystyle =\frac{3}{10}
\displaystyle P(X=3)
\displaystyle =P(\text{3\ red\ balls})
\displaystyle =\frac{{}^{4}C_{3}}{{}^{10}C_{3}}
\displaystyle =\frac{4}{120}
\displaystyle =\frac{1}{30}
\displaystyle \text{Thus,\ the\ probability\ distribution\ of\ }X\ \text{is\ given\ by}
\displaystyle \begin{array}{c|cccc} X & 0 & 1 & 2 & 3 \\ \hline P(X) & \frac{1}{6} & \frac{1}{2} & \frac{3}{10} & \frac{1}{30} \end{array}

\displaystyle \textbf{Question 11: }~\text{Five defective mangoes are accidentally mixed with 15 good ones. Four} \\ \text{mangoes are drawn at random from this lot. Find the probability distribution of the} \\ \text{number of defective mangoes.}
\displaystyle \text{Answer:}
\displaystyle \text{Let } X \text{ denote the number of defective mangoes in a sample of } 4 \text{ mangoes drawn} \\ \text{from a bag containing } 5 \text{ defective mangoes and } 15 \text{ good mangoes. Then } \\ X \text{ can take the values } 0, 1, 2, 3 \text{ and } 4.
\displaystyle \text{Now,}
\displaystyle P(X=0)
\displaystyle =P(\text{no\ defective\ mango})
\displaystyle =\frac{{}^{15}C_{4}}{{}^{20}C_{4}}
\displaystyle =\frac{1365}{4845}
\displaystyle =\frac{91}{323}
\displaystyle P(X=1)
\displaystyle =P(\text{1\ defective\ mango})
\displaystyle =\frac{{}^{5}C_{1}\times{}^{15}C_{3}}{{}^{20}C_{4}}
\displaystyle =\frac{2275}{4845}
\displaystyle =\frac{455}{969}
\displaystyle P(X=2)
\displaystyle =P(\text{2\ defective\ mangoes})
\displaystyle =\frac{{}^{5}C_{2}\times{}^{15}C_{2}}{{}^{20}C_{4}}
\displaystyle =\frac{1050}{4845}
\displaystyle =\frac{70}{323}
\displaystyle P(X=3)
\displaystyle =P(\text{3\ defective\ mangoes})
\displaystyle =\frac{{}^{5}C_{3}\times{}^{15}C_{1}}{{}^{20}C_{4}}
\displaystyle =\frac{150}{4845}
\displaystyle =\frac{10}{323}
\displaystyle P(X=4)
\displaystyle =P(\text{4\ defective\ mangoes})
\displaystyle =\frac{{}^{5}C_{4}}{{}^{20}C_{4}}
\displaystyle =\frac{5}{4845}
\displaystyle =\frac{1}{969}
\displaystyle \text{Thus,\ the\ probability\ distribution\ of\ }X\ \text{is\ given\ by}
\displaystyle \begin{array}{c|ccccc} X & 0 & 1 & 2 & 3 & 4 \\ \hline P(X) & \frac{91}{323} & \frac{455}{969} & \frac{70}{323} & \frac{10}{323} & \frac{1}{969} \end{array}

\displaystyle \textbf{Question 12: }~\text{Two dice are thrown together and the number appearing on them noted. } \\ X\text{ denotes the sum of the two numbers. Assuming that all the 36 outcomes are equally likely,} \\ \text{what is the probability distribution of }X\text{?}
\displaystyle \text{Answer:}
\displaystyle \text{Let } X \text{ denote the sum of the numbers on two dice. Then } X \text{ can take the values } \\ 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 \text{ and } 12.
\displaystyle \text{Sample\ space:}\ \{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,1), \\ (3,2),(3,3),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),(5,1),(5,2), \\ (5,3),(5,4),(5,5),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\}
\displaystyle \text{Now,}
\displaystyle P(X=2)=\frac{1}{36}
\displaystyle P(X=3)=\frac{2}{36}
\displaystyle P(X=4)=\frac{3}{36}
\displaystyle P(X=5)=\frac{4}{36}
\displaystyle P(X=6)=\frac{5}{36}
\displaystyle P(X=7)=\frac{6}{36}
\displaystyle P(X=8)=\frac{5}{36}
\displaystyle P(X=9)=\frac{4}{36}
\displaystyle P(X=10)=\frac{3}{36}
\displaystyle P(X=11)=\frac{2}{36}
\displaystyle P(X=12)=\frac{1}{36}
\displaystyle \text{Thus,\ the\ probability\ distribution\ of\ }X\ \text{is\ given\ by}
\displaystyle \begin{array}{c|ccccccccccc} X & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ \hline P(X) & \frac{1}{36} & \frac{2}{36} & \frac{3}{36} & \frac{4}{36} & \frac{5}{36} & \frac{6}{36} & \frac{5}{36} & \frac{4}{36} & \frac{3}{36} & \frac{2}{36} & \frac{1}{36} \end{array}

\displaystyle \textbf{Question 13: }~\text{A class has 15 students whose ages are }14,17,15,14,21,19,20,16,18,17,20 \\ 17,16,19 \text{ and }20\text{ years respectively. One student is selected in such a manner that each has} \\ \text{the same chance of being selected and the age }X\text{ of the selected student is recorded. What} \\ \text{is the probability distribution of the random variable }X\text{?}
\displaystyle \text{Answer:}
\displaystyle \text{Here,}\ X\ \text{can\ take\ the\ values\ }14,15,16,17,18,19,20\ \text{and\ }21.
\displaystyle \text{Now,}
\displaystyle P(X=14)=\frac{2}{15}
\displaystyle P(X=15)=\frac{1}{15}
\displaystyle P(X=16)=\frac{2}{15}
\displaystyle P(X=17)=\frac{3}{15}
\displaystyle P(X=18)=\frac{1}{15}
\displaystyle P(X=19)=\frac{2}{15}
\displaystyle P(X=20)=\frac{3}{15}
\displaystyle P(X=21)=\frac{1}{15}
\displaystyle \text{Thus,\ the\ probability\ distribution\ of\ }X\ \text{is\ given\ by}
\displaystyle \begin{array}{c|cccccccc} X & 14 & 15 & 16 & 17 & 18 & 19 & 20 & 21 \\ \hline P(X) & \frac{2}{15} & \frac{1}{15} & \frac{2}{15} & \frac{3}{15} & \frac{1}{15} & \frac{2}{15} & \frac{3}{15} & \frac{1}{15} \end{array}

\displaystyle \textbf{Question 14: }~\text{Five defective bolts are accidentally mixed with twenty good ones. If four} \\ \text{bolts are drawn at random from this lot, find the probability distribution of the number of} \\ \text{defective bolts.}
\displaystyle \text{Answer:}
\displaystyle \text{Let } X \text{ denote the number of defective bolts in a sample of } 4 \text{ bolts drawn from a} \\ \text{bag containing } 5 \text{ defective bolts and } 20 \text{ good bolts. Then } X \text{ can take the values } \\ 0, 1, 2, 3 \text{ and } 4.
\displaystyle \text{Now,}
\displaystyle P(X=0)
\displaystyle =P(\text{no\ defective\ bolts})
\displaystyle =\frac{{}^{20}C_{4}}{{}^{25}C_{4}}
\displaystyle =\frac{4845}{12650}
\displaystyle =\frac{969}{2530}
\displaystyle P(X=1)
\displaystyle =P(\text{1\ defective\ bolt})
\displaystyle =\frac{{}^{5}C_{1}\times{}^{20}C_{3}}{{}^{25}C_{4}}
\displaystyle =\frac{5700}{12650}
\displaystyle =\frac{114}{253}
\displaystyle P(X=2)
\displaystyle =P(\text{2\ defective\ bolts})
\displaystyle =\frac{{}^{5}C_{2}\times{}^{20}C_{2}}{{}^{25}C_{4}}
\displaystyle =\frac{1900}{12650}
\displaystyle =\frac{38}{253}
\displaystyle P(X=3)
\displaystyle =P(\text{3\ defective\ bolts})
\displaystyle =\frac{{}^{5}C_{3}\times{}^{20}C_{1}}{{}^{25}C_{4}}
\displaystyle =\frac{200}{12650}
\displaystyle =\frac{4}{253}
\displaystyle P(X=4)
\displaystyle =P(\text{4\ defective\ bolts})
\displaystyle =\frac{{}^{5}C_{4}}{{}^{25}C_{4}}
\displaystyle =\frac{5}{12650}
\displaystyle =\frac{1}{2530}
\displaystyle \text{Thus,\ the\ probability\ distribution\ of\ }X\ \text{is\ given\ by}
\displaystyle \begin{array}{c|ccccc} X & 0 & 1 & 2 & 3 & 4 \\ \hline P(X) & \frac{969}{2530} & \frac{114}{253} & \frac{38}{253} & \frac{4}{253} & \frac{1}{2530} \end{array}

\displaystyle \textbf{Question 15: }~\text{Two cards are drawn successively with replacement from a well shuffled pack} \\ \text{of 52 cards. Find the probability distribution of the number of aces.}
\displaystyle \text{Answer:}
\displaystyle \text{Let } X \text{ denote the number of aces in a sample of } 2 \text{ cards drawn from a well-shuffled pack of } \\ 52 \text{ playing cards. Then } X \text{ can take the values } 0, 1 \text{ and } 2.
\displaystyle \text{Now,}
\displaystyle P(X=0)
\displaystyle =P(\text{no\ ace})
\displaystyle =\frac{48}{52}\times\frac{48}{52}
\displaystyle =\frac{12}{13}\times\frac{12}{13}
\displaystyle =\frac{144}{169}
\displaystyle P(X=1)
\displaystyle =P(\text{1\ ace})
\displaystyle =\frac{4}{52}\times\frac{48}{52}+\frac{48}{52}\times\frac{4}{52}
\displaystyle =\frac{2}{13}\times\frac{12}{13}
\displaystyle =\frac{24}{169}
\displaystyle P(X=2)
\displaystyle =P(\text{2\ aces})
\displaystyle =\frac{4}{52}\times\frac{4}{52}
\displaystyle =\frac{1}{13}\times\frac{1}{13}
\displaystyle =\frac{1}{169}
\displaystyle \text{Thus,\ the\ probability\ distribution\ of\ }X\ \text{is\ given\ by}
\displaystyle \begin{array}{c|ccc} X & 0 & 1 & 2 \\ \hline P(X) & \frac{144}{169} & \frac{24}{169} & \frac{1}{169} \end{array}

\displaystyle \textbf{Question 16: }~\text{Two cards are drawn successively with replacement from a well shuffled pack} \\ \text{of 52 cards. Find the probability distribution of the number of kings.}
\displaystyle \text{Answer:}
\displaystyle \text{Let } X \text{ denote the number of kings in a sample of } 2 \text{ cards drawn from a well-shuffled pack of } \\  52 \text{ playing cards. Then } X \text{ can take the values } 0, 1 \text{ and } 2.
\displaystyle \text{Now,}
\displaystyle P(X=0)
\displaystyle =P(\text{no\ kings})
\displaystyle =\frac{48}{52}\times\frac{48}{52}
\displaystyle =\frac{12}{13}\times\frac{12}{13}
\displaystyle =\frac{144}{169}
\displaystyle P(X=1)
\displaystyle =P(\text{1\ king})
\displaystyle =\frac{4}{52}\times\frac{48}{52}+\frac{48}{52}\times\frac{4}{52}
\displaystyle =\frac{2}{13}\times\frac{12}{13}
\displaystyle =\frac{24}{169}
\displaystyle P(X=2)
\displaystyle =P(\text{2\ kings})
\displaystyle =\frac{4}{52}\times\frac{4}{52}
\displaystyle =\frac{1}{13}\times\frac{1}{13}
\displaystyle =\frac{1}{169}
\displaystyle \text{Thus,\ the\ probability\ distribution\ of\ }X\ \text{is\ given\ by}
\displaystyle \begin{array}{c|ccc} X & 0 & 1 & 2 \\ \hline P(X) & \frac{144}{169} & \frac{24}{169} & \frac{1}{169} \end{array}

\displaystyle \textbf{Question 17: }~\text{Two cards are drawn successively without replacement from a well shuffled pack} \\ \text{of 52 cards. Find the probability distribution of the number of aces.}
\displaystyle \text{Answer:}
\displaystyle \text{Let } X \text{ denote the number of aces in a sample of } 2 \text{ cards drawn from a well-shuffled pack of } \\ 52 \text{ playing cards. Then } X \text{ can take the values } 0, 1 \text{ and } 2.
\displaystyle \text{Now,}
\displaystyle P(X=0)
\displaystyle =P(\text{no\ ace})
\displaystyle =\frac{48}{52}\times\frac{47}{51}
\displaystyle =\frac{2256}{2652}
\displaystyle =\frac{188}{221}
\displaystyle P(X=1)
\displaystyle =P(\text{1\ ace})
\displaystyle =\frac{4}{52}\times\frac{48}{51}+\frac{48}{52}\times\frac{4}{51}
\displaystyle =\frac{384}{2652}
\displaystyle =\frac{32}{221}
\displaystyle P(X=2)
\displaystyle =P(\text{2\ aces})
\displaystyle =\frac{4}{52}\times\frac{3}{51}
\displaystyle =\frac{12}{2652}
\displaystyle =\frac{1}{221}
\displaystyle \text{Thus,\ the\ probability\ distribution\ of\ }X\ \text{is\ given\ by}
\displaystyle \begin{array}{c|ccc} X & 0 & 1 & 2 \\ \hline P(X) & \frac{188}{221} & \frac{32}{221} & \frac{1}{221} \end{array}

\displaystyle \textbf{Question 18: }~\text{Find the probability distribution of the number of white balls drawn in a random} \\ \text{draw of 3 balls without replacement, from a bag containing 4 white and 6 red balls.}
\displaystyle \text{Answer:}
\displaystyle \text{Let } X \text{ denote the number of white balls in a sample of } 3 \text{ balls drawn from a bag containing } \\ 4 \text{ white and } 6 \text{ red balls. Then } X \text{ can take the values } 0, 1, 2 \text{ and } 3.
\displaystyle \text{Now,}
\displaystyle P(X=0)
\displaystyle =P(\text{no\ white\ ball})
\displaystyle =\frac{{}^{6}C_{3}}{{}^{10}C_{3}}
\displaystyle =\frac{20}{120}
\displaystyle =\frac{1}{6}
\displaystyle P(X=1)
\displaystyle =P(\text{1\ white\ ball})
\displaystyle =\frac{{}^{4}C_{1}\times{}^{6}C_{2}}{{}^{10}C_{3}}
\displaystyle =\frac{60}{120}
\displaystyle =\frac{1}{2}
\displaystyle P(X=2)
\displaystyle =P(\text{2\ white\ balls})
\displaystyle =\frac{{}^{4}C_{2}\times{}^{6}C_{1}}{{}^{10}C_{3}}
\displaystyle =\frac{36}{120}
\displaystyle =\frac{3}{10}
\displaystyle P(X=3)
\displaystyle =P(\text{3\ white\ balls})
\displaystyle =\frac{{}^{4}C_{3}}{{}^{10}C_{3}}
\displaystyle =\frac{4}{120}
\displaystyle =\frac{1}{30}
\displaystyle \text{Thus,\ the\ probability\ distribution\ of\ }X\ \text{is\ given\ by}
\displaystyle \begin{array}{c|cccc} X & 0 & 1 & 2 & 3 \\ \hline P(X) & \frac{1}{6} & \frac{1}{2} & \frac{3}{10} & \frac{1}{30} \end{array}

\displaystyle \textbf{Question 19: }~\text{Find the probability distribution of }Y\text{ in two throws of two dice, where } \\ Y\text{ represents the number of times a total of }9\text{ appears.}
\displaystyle \text{Answer:}
\displaystyle \text{It is given that } Y \text{ denotes the number of times a total of } 9 \text{ appears on throwing} \\ \text{a pair of dice.}
\displaystyle \text{When the dice are thrown } 2 \text{ times, the combinations giving a total of } 9 \text{ are}
\displaystyle (3,6),(4,5),(5,4),(6,3)
\displaystyle \text{So,\ the\ total\ number\ of\ outcomes\ is}\ 36\ \text{and\ the\ total\ number\ of\ favourable\ outcomes\ is}\ 4.
\displaystyle \text{Probability\ of\ getting\ a\ total\ of}\ 9=\frac{4}{36}=\frac{1}{9}
\displaystyle \text{Probability\ of\ not\ getting\ a\ total\ of}\ 9=1-\frac{1}{9}=\frac{8}{9}
\displaystyle \text{If}\ Y\ \text{takes\ the\ values}\ 0,1,2,\ \text{then}
\displaystyle P(Y=0)=\frac{8}{9}\times\frac{8}{9}=\frac{64}{81}
\displaystyle P(Y=1)=\frac{1}{9}\times\frac{8}{9}+\frac{8}{9}\times\frac{1}{9}=\frac{16}{81}
\displaystyle P(Y=2)=\frac{1}{9}\times\frac{1}{9}=\frac{1}{81}
\displaystyle \text{Thus,\ the\ probability\ distribution\ of\ }Y\ \text{is\ given\ by}
\displaystyle \begin{array}{c|ccc} Y & 0 & 1 & 2 \\ \hline P(Y) & \frac{64}{81} & \frac{16}{81} & \frac{1}{81} \end{array}

\displaystyle \textbf{Question 20: }~\text{From a lot containing 25 items, 5 of which are defective, 4 are chosen at random.} \\ \text{Let }X\text{ be the number of defectives found. Obtain the probability distribution of } \\ X\text{ if the items are chosen without replacement.}
\displaystyle \text{Answer:}
\displaystyle \text{Let } X \text{ denote the number of defective items in a sample of } 4 \text{ items drawn from a bag containing } \\ 5 \text{ defective items and } 20 \text{ good items. Then } X \text{ can take the values } 0, 1, 2, 3 \text{ and } 4.
\displaystyle \text{Now,}
\displaystyle P(X=0)
\displaystyle =P(\text{no\ defective\ item})
\displaystyle =\frac{{}^{20}C_{4}}{{}^{25}C_{4}}
\displaystyle =\frac{4845}{12650}
\displaystyle =\frac{969}{2530}
\displaystyle P(X=1)
\displaystyle =P(\text{1\ defective\ item})
\displaystyle =\frac{{}^{5}C_{1}\times{}^{20}C_{3}}{{}^{25}C_{4}}
\displaystyle =\frac{5700}{12650}
\displaystyle =\frac{114}{253}
\displaystyle P(X=2)
\displaystyle =P(\text{2\ defective\ items})
\displaystyle =\frac{{}^{5}C_{2}\times{}^{20}C_{2}}{{}^{25}C_{4}}
\displaystyle =\frac{1900}{12650}
\displaystyle =\frac{38}{253}
\displaystyle P(X=3)
\displaystyle =P(\text{3\ defective\ items})
\displaystyle =\frac{{}^{5}C_{3}\times{}^{20}C_{1}}{{}^{25}C_{4}}
\displaystyle =\frac{200}{12650}
\displaystyle =\frac{4}{253}
\displaystyle P(X=4)
\displaystyle =P(\text{4\ defective\ items})
\displaystyle =\frac{{}^{5}C_{4}}{{}^{25}C_{4}}
\displaystyle =\frac{5}{12650}
\displaystyle =\frac{1}{2530}
\displaystyle \text{Thus,\ the\ probability\ distribution\ of\ }X\ \text{is\ given\ by}
\displaystyle \begin{array}{c|ccccc} X & 0 & 1 & 2 & 3 & 4 \\ \hline P(X) & \frac{969}{2530} & \frac{114}{253} & \frac{38}{253} & \frac{4}{253} & \frac{1}{2530} \end{array}

\displaystyle \textbf{Question 21: }~\text{Three cards are drawn successively with replacement from a well shuffled deck} \\ \text{of 52 cards. A random variable }X\text{ denotes the number of hearts in the three cards} \\ \text{drawn. Determine the probability distribution of }X\text{.}
\displaystyle \text{Answer:}
\displaystyle \text{Let } X \text{ denote the number of hearts in a sample of } 3 \text{ cards drawn from a well-shuffled deck of }\\  52 \text{ cards. Then } X \text{ can take the values } 0, 1, 2 \text{ and } 3.
\displaystyle \text{Now,}
\displaystyle P(X=0)
\displaystyle =P(\text{no\ heart})
\displaystyle =\frac{39}{52}\times\frac{39}{52}\times\frac{39}{52}
\displaystyle =\frac{27}{64}
\displaystyle P(X=1)
\displaystyle =P(\text{1\ heart})
\displaystyle =\frac{13}{52}\times\frac{39}{52}\times\frac{39}{52}+\frac{39}{52}\times\frac{13}{52}\times\frac{39}{52}+\frac{39}{52}\times\frac{39}{52}\times\frac{13}{52}
\displaystyle =\frac{27}{64}
\displaystyle P(X=2)
\displaystyle =P(\text{2\ hearts})
\displaystyle =\frac{13}{52}\times\frac{13}{52}\times\frac{39}{52}+\frac{39}{52}\times\frac{13}{52}\times\frac{13}{52}+\frac{13}{52}\times\frac{39}{52}\times\frac{13}{52}
\displaystyle =\frac{9}{64}
\displaystyle P(X=3)
\displaystyle =P(\text{3\ hearts})
\displaystyle =\frac{13}{52}\times\frac{13}{52}\times\frac{13}{52}
\displaystyle =\frac{1}{64}
\displaystyle \text{Thus,\ the\ probability\ distribution\ of\ }X\ \text{is\ given\ by}
\displaystyle \begin{array}{c|cccc} X & 0 & 1 & 2 & 3 \\ \hline P(X) & \frac{27}{64} & \frac{27}{64} & \frac{9}{64} & \frac{1}{64} \end{array}

\displaystyle \textbf{Question 22: }~\text{An urn contains 4 red and 3 blue balls. Find the probability distribution of} \\ \text{the number of blue balls in a random draw of 3 balls with replacement.}
\displaystyle \text{Answer:}
\displaystyle \text{Let } X \text{ denote the number of blue balls in a sample of } 3 \text{ balls drawn from a bag containing } \\  4 \text{ red and } 3 \text{ blue balls. Then } X \text{ can take the values } 0, 1, 2 \text{ and } 3.
\displaystyle \text{Now,}
\displaystyle P(X=0)=P(\text{no\ blue\ ball})=\frac{4}{7}\times\frac{4}{7}\times\frac{4}{7}=\frac{64}{343}
\displaystyle P(X=1)=P(\text{1\ blue\ ball})=\frac{3}{7}\times\frac{4}{7}\times\frac{4}{7}+\frac{4}{7}\times\frac{3}{7}\times\frac{4}{7}+\frac{4}{7}\times\frac{4}{7}\times\frac{3}{7}=\frac{144}{343}
\displaystyle P(X=2)=P(\text{2\ blue\ balls})=\frac{3}{7}\times\frac{3}{7}\times\frac{4}{7}+\frac{4}{7}\times\frac{3}{7}\times\frac{3}{7}+\frac{3}{7}\times\frac{4}{7}\times\frac{3}{7}=\frac{108}{343}
\displaystyle P(X=3)=P(\text{3\ blue\ balls})=\frac{3}{7}\times\frac{3}{7}\times\frac{3}{7}=\frac{27}{343}
\displaystyle \text{Thus,\ the\ probability\ distribution\ of\ }X\ \text{is\ given\ by}
\displaystyle \begin{array}{c|cccc} X & 0 & 1 & 2 & 3 \\ \hline P(X) & \frac{64}{343} & \frac{144}{343} & \frac{108}{343} & \frac{27}{343} \end{array}

\displaystyle \textbf{Question 23: }~\text{Two cards are drawn simultaneously from a well shuffled deck of 52 cards.} \\ \text{Find the probability distribution of the number of successes, when getting a spade is considered} \\ \text{a success.}
\displaystyle \text{Answer:}
\displaystyle \text{Let } X \text{ denote the number of blue balls in a sample of } 3 \text{ balls drawn from a bag containing } \\ 4 \text{ red and } 3 \text{ blue balls. Then } X \text{ can take the values } 0, 1, 2 \text{ and } 3.
\displaystyle \text{Now,}
\displaystyle P(X=0)
\displaystyle =P(\text{no\ spade})
\displaystyle =\frac{{}^{39}C_{2}}{{}^{52}C_{2}}
\displaystyle =\frac{741}{1326}
\displaystyle =\frac{19}{34}
\displaystyle P(X=1)
\displaystyle =P(\text{1\ spade})
\displaystyle =\frac{{}^{13}C_{1}\times{}^{39}C_{1}}{{}^{52}C_{2}}
\displaystyle =\frac{507}{1326}
\displaystyle =\frac{13}{34}
\displaystyle P(X=2)
\displaystyle =P(\text{2\ spades})
\displaystyle =\frac{{}^{13}C_{2}}{{}^{52}C_{2}}
\displaystyle =\frac{78}{1326}
\displaystyle =\frac{1}{17}
\displaystyle \text{Thus,\ the\ probability\ distribution\ of\ }X\ \text{is\ given\ by}
\displaystyle \begin{array}{c|ccc} X & 0 & 1 & 2 \\ \hline P(X) & \frac{19}{34} & \frac{13}{34} & \frac{1}{17} \end{array}

\displaystyle \textbf{Question 24: }~\text{A fair die is tossed twice. If the number appearing on the top is less than } \\ 3,\text{ it is a success. Find the probability distribution of number of successes.}
\displaystyle \text{Answer:}
\displaystyle \text{Let } X \text{ denote the number of blue balls in a sample of } 3 \text{ balls drawn from a bag containing } \\ 4 \text{ red and } 3 \text{ blue balls. Then } X \text{ can take the values } 0, 1, 2 \text{ and } 3.
\displaystyle \text{Now,}
\displaystyle P(X=0)=\frac{16}{36}=\frac{4}{9}
\displaystyle P(X=1)=\frac{16}{36}=\frac{4}{9}
\displaystyle P(X=2)=\frac{4}{36}=\frac{1}{9}
\displaystyle \text{Thus,\ the\ probability\ distribution\ of\ }X\ \text{is\ given\ by}
\displaystyle \begin{array}{c|ccc} X & 0 & 1 & 2 \\ \hline P(X) & \frac{4}{9} & \frac{4}{9} & \frac{1}{9} \end{array}

\displaystyle \textbf{Question 25: }~\text{An urn contains 5 red and 2 black balls. Two balls are randomly selected.} \\ \text{Let }X\text{ represent the number of black balls. What are the possible values of }X\text{? Is }X\text{ a} \\ \text{random variable?}
\displaystyle \text{Answer:}
\displaystyle \text{The possible values of } X \text{ are } 0, 1 \text{ and } 2, \text{ i.e. no black ball, } 1 \text{ black ball and }\\  2 \text{ black balls.}
\displaystyle \text{Yes, } X \text{ is a random variable.}
\displaystyle \text{A random variable is a real-valued function having the sample space associated with a} \\ \text{random experiment as its domain.}

\displaystyle \textbf{Question 26: }~\text{Let }X\text{ represent the difference between the number of heads and the} \\ \text{number of tails when a coin is tossed 6 times. What are possible values of }X\text{?}
\displaystyle \text{Answer:}
\displaystyle \text{A coin is tossed } 6 \text{ times and } X \text{ represents the difference between the number of heads} \\ \text{and the number of tails.}
\displaystyle \text{Sample\ space\ of\ the\ experiment\ is}
\displaystyle S=\{(0\ \text{heads},6\ \text{tails}),(1\ \text{head},5\ \text{tails}),(2\ \text{heads},4\ \text{tails}), \\ (3\ \text{heads},3\ \text{tails}),(4\ \text{heads},2\ \text{tails}),(5\ \text{heads},1\ \text{tail}),(6\ \text{heads},0\ \text{tails})\}
\displaystyle \text{The\ values\ of}\ X\ \text{corresponding\ to\ these\ outcomes\ are\ as\ follows:}
\displaystyle \therefore\ X(0\ \text{heads},6\ \text{tails})=0-6=-6
\displaystyle X(1\ \text{head},5\ \text{tails})=1-5=-4
\displaystyle X(2\ \text{heads},4\ \text{tails})=2-4=-2
\displaystyle X(3\ \text{heads},3\ \text{tails})=3-3=0
\displaystyle X(4\ \text{heads},2\ \text{tails})=4-2=2
\displaystyle X(5\ \text{heads},1\ \text{tail})=5-1=4
\displaystyle X(6\ \text{heads},0\ \text{tails})=6-0=6
\displaystyle \therefore\ \text{Possible\ values\ of}\ X\ \text{are}\ \{-6,-4,-2,0,2,4,6\}.

\displaystyle \textbf{Question 27: }~\text{From a lot of 10 bulbs, which includes 3 defectives, a sample of 2 bulbs is} \\ \text{drawn at random. Find the probability distribution of the number of defective bulbs.}
\displaystyle \text{Answer:}
$\displaystyle \text{Let } X \text{ denote the number of defective bulbs in a sample of } 2 \text{ bulbs drawn from a lot of }\\  10 \text{ bulbs containing } 3 \text{ defectives and } 7 \text{ non-defectives. Then } X \text{ can take the values } \\ 0, 1 \text{ and } 2.
\displaystyle \text{Now,}
\displaystyle P(X=0)=P(\text{no\ defective\ bulb})=\frac{{}^{7}C_{2}}{{}^{10}C_{2}}=\frac{7}{15}
\displaystyle P(X=1)=P(\text{1\ defective\ bulb})=\frac{{}^{3}C_{1}\times{}^{7}C_{1}}{{}^{10}C_{2}}=\frac{7}{15}
\displaystyle P(X=2)=P(\text{2\ defective\ bulbs})=\frac{{}^{3}C_{2}}{{}^{10}C_{2}}=\frac{1}{15}
\displaystyle \text{Thus,\ the\ probability\ distribution\ of\ }X\ \text{is\ given\ below,}
\displaystyle \begin{array}{c|ccc} X & 0 & 1 & 2 \\ \hline P(X) & \frac{7}{15} & \frac{7}{15} & \frac{1}{15} \end{array}

\displaystyle \textbf{Question 28: }~\text{Four balls are to be drawn without replacement from a box containing 8 red} \\ \text{and 4 white balls. If }X\text{ denotes the number of red balls drawn, find the probability distribution of }X\text{.}
\displaystyle \text{Answer:}
\displaystyle \text{As four balls are drawn without replacement and } X \text{ denotes the number of red balls drawn,}
\displaystyle \text{so } X \text{ is a random variable that can take the values } 0, 1, 2, 3 \text{ or } 4.
\displaystyle \text{Now,}
\displaystyle P(X=0)=P(\text{all\ white\ balls})=P(\text{WWWW})=\frac{4}{12}\times\frac{3}{11}\times\frac{2}{10}\times\frac{1}{9}=\frac{1}{495}
\displaystyle P(X=1)=P(\text{one\ red\ and\ three\ white\ balls})
\displaystyle =P(\text{RWWW})+P(\text{WRWW})+P(\text{WWRW})+P(\text{WWWR})
\displaystyle =\frac{8}{12}\times\frac{4}{11}\times\frac{3}{10}\times\frac{2}{9}+\frac{4}{12}\times\frac{8}{11}\times\frac{3}{10}\times\frac{2}{9}+\frac{4}{12}\times\frac{3}{11}\times\frac{8}{10}\times\frac{2}{9}+\frac{4}{12}\times\frac{3}{11}\times\frac{2}{10}\times\frac{8}{9}
\displaystyle =4\times\frac{8}{495}=\frac{32}{495}
\displaystyle P(X=2)=P(\text{two\ red\ and\ two\ white\ balls})
\displaystyle =P(\text{RRWW})+P(\text{RWRW})+P(\text{RWWR})+P(\text{WWRR})+P(\text{WRWR})+P(\text{WRRW})
\displaystyle =\frac{8}{12}\times\frac{7}{11}\times\frac{4}{10}\times\frac{3}{9}+\frac{8}{12}\times\frac{4}{11}\times\frac{7}{10}\times\frac{3}{9}+\frac{8}{12}\times\frac{4}{11}\times\frac{3}{10}\times\frac{7}{9}+\frac{4}{12}\times\frac{3}{11}\times\frac{8}{10}\times\frac{7}{9}+\frac{4}{12}\times\frac{8}{11}\times\frac{3}{10}\times\frac{7}{9}+\frac{4}{12}\times\frac{7}{11}\times\frac{8}{10}\times\frac{3}{9}
\displaystyle =6\times\frac{28}{495}=\frac{168}{495}
\displaystyle P(X=3)=P(\text{three\ red\ and\ one\ white\ ball})
\displaystyle =P(\text{RRRW})+P(\text{RRWR})+P(\text{RWRR})+P(\text{WRRR})
\displaystyle =\frac{8}{12}\times\frac{7}{11}\times\frac{6}{10}\times\frac{4}{9}+\frac{8}{12}\times\frac{7}{11}\times\frac{4}{10}\times\frac{6}{9}+\frac{8}{12}\times\frac{4}{11}\times\frac{7}{10}\times\frac{6}{9}+\frac{4}{12}\times\frac{8}{11}\times\frac{7}{10}\times\frac{6}{9}
\displaystyle =4\times\frac{56}{495}=\frac{224}{495}
\displaystyle P(X=4)=P(\text{all\ red\ balls})=P(\text{RRRR})=\frac{8}{12}\times\frac{7}{11}\times\frac{6}{10}\times\frac{5}{9}=\frac{70}{495}
\displaystyle \text{So,\ the\ probability\ distribution\ of}\ X\ \text{is\ as\ follows:}
\displaystyle X\qquad 0\qquad 1\qquad 2\qquad 3\qquad 4
\displaystyle P(X)\qquad \frac{1}{495}\qquad \frac{32}{495}\qquad \frac{168}{495}\qquad \frac{224}{495}\qquad \frac{70}{495}

\displaystyle \textbf{Question 29: }~\text{The probability distribution of a random variable }X \text{ is given below:}
\displaystyle \begin{array}{c|cccc} X & 0 & 1 & 2 & 3\\ \hline P(X) & k & \frac{k}{2} & \frac{k}{4} & \frac{k}{8} \end{array}
\displaystyle \text{(i) Determine the value of }k\ \text{(ii) Determine }P(X\le 2)\text{ and }P(X>2)\ \text{(iii) Find }P(X\le 2)+P(X>2).
\displaystyle \text{Answer:}
\displaystyle \text{(i)  }
\displaystyle \text{We\ have,}
\displaystyle \text{The\ probability\ distribution\ of\ a\ random\ variable}\ X\ \text{is\ given\ below:}
\displaystyle x\qquad 0\qquad 1\qquad 2\qquad 3
\displaystyle P(X)\qquad k\qquad \frac{k}{2}\qquad \frac{k}{4}\qquad \frac{k}{8}
\displaystyle \text{As,}\ \sum p_{i}=1
\displaystyle \Rightarrow k+\frac{k}{2}+\frac{k}{4}+\frac{k}{8}=1
\displaystyle \Rightarrow \frac{8k+4k+2k+k}{8}=1
\displaystyle \Rightarrow \frac{15k}{8}=1
\displaystyle \therefore\ k=\frac{8}{15}
\displaystyle \text{(ii)  }
\displaystyle \text{We\ have,}
\displaystyle \text{The\ probability\ distribution\ of\ a\ random\ variable}\ X\ \text{is\ given\ below:}
\displaystyle x\qquad 0\qquad 1\qquad 2\qquad 3
\displaystyle P(X)\qquad k\qquad \frac{k}{2}\qquad \frac{k}{4}\qquad \frac{k}{8}
\displaystyle \text{As,}\ P(X\le 2)=1-P(X=3)
\displaystyle =1-\frac{k}{8}
\displaystyle =1-\frac{8}{15\times 8}
\displaystyle =1-\frac{1}{15}
\displaystyle =\frac{14}{15}
\displaystyle \text{Also,}\ P(X>2)=P(X=3)
\displaystyle =\frac{8}{15\times 8}
\displaystyle =\frac{1}{15}
\displaystyle \text{(iii)  }
\displaystyle \text{We\ have,}
\displaystyle \text{The\ probability\ distribution\ of\ a\ random\ variable}\ X\ \text{is\ given\ below:}
\displaystyle x\qquad 0\qquad 1\qquad 2\qquad 3
\displaystyle P(X)\qquad k\qquad \frac{k}{2}\qquad \frac{k}{4}\qquad \frac{k}{8}
\displaystyle P(X\le 2)+P(X>2)
\displaystyle =\frac{14}{15}+\frac{1}{15}\ \text{Using\ (ii)}
\displaystyle =\frac{15}{15}
\displaystyle =1

\displaystyle \textbf{Question 30: }~\text{Let }X\text{ denote the number of colleges where you will apply after your} \\ \text{results and }P(X=x)\text{ denotes your probability of getting admission in }x\text{ number of colleges.} \\ \text{It is given that}
\displaystyle P(X=x)=\begin{cases} kx,\ \text{if }x=0,1\ 2kx,\ \text{if }x=2\ k(5-x),\ \text{if }x=3\text{ or }4\ 0,\ \text{if }x>4 \end{cases}
\displaystyle \text{where }k\text{ is a positive constant. Find the value of }k.\ \text{Also find the probability that} \\ \text{you will get admission in (i) exactly one college (ii) at most 2 colleges (iii) at least 2 colleges.}
\displaystyle \text{Answer:}
\displaystyle \text{The\ probability\ distribution\ of}\ X\ \text{is}
\displaystyle X\qquad 0\qquad 1\qquad 2\qquad 3\qquad 4
\displaystyle P(X)\qquad 0\qquad k\qquad 4k\qquad 2k\qquad k
\displaystyle \text{The\ given\ distribution\ is\ a\ probability\ distribution.}
\displaystyle \therefore\ \sum p_{i}=1
\displaystyle \Rightarrow 0+k+4k+2k+k=1
\displaystyle \Rightarrow 8k=1
\displaystyle \Rightarrow k=0.125
\displaystyle \text{(i)}\ P(\text{getting\ admission\ in\ exactly\ one\ college})=P(X=1)=k=0.125
\displaystyle \text{(ii)}\ P(\text{getting\ admission\ in\ at\ most\ }2\ \text{colleges})=P(X\le 2)=0+k+4k=5k=0.625
\displaystyle \text{(iii)}\ P(\text{getting\ admission\ in\ at\ least\ }2\ \text{colleges})=P(X\ge 2)=4k+2k+k=7k=0.875

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