\displaystyle \textbf{Question 1: }~\text{From a pack of }52\text{ cards, two are drawn one by one without replacement.} \\ \text{Find the probability that both of them are kings.}
\displaystyle \text{Answer:}
\displaystyle \text{Consider the given events.}
\displaystyle A=\text{A king in the first draw}
\displaystyle B=\text{A king in the second draw}
\displaystyle \text{Now,}
\displaystyle P(A)=\frac{4}{52}=\frac{1}{13}
\displaystyle P(B\mid A)=\text{Getting a king in the second draw after getting a king in the first draw}
\displaystyle =\frac{3}{51}\ \text{[After the first draw, the total number of cards will be }51.\text{ Then, }3\text{ kings will be remaining.]}
\displaystyle =\frac{1}{17}
\displaystyle \therefore\ \text{Required probability}=P(A\cap B)=P(A)\times P(B\mid A)=\frac{1}{13}\times\frac{1}{17}=\frac{1}{221}

\displaystyle \textbf{Question 2: }~\text{From a pack of }52\text{ cards, }4\text{ are drawn one by one without replacement.} \\ \text{Find the probability that all are aces (or, kings). }
\displaystyle \text{Answer:}
\displaystyle \text{Consider the given events.}
\displaystyle A=\text{An ace in the first draw}
\displaystyle B=\text{An ace in the second draw}
\displaystyle C=\text{An ace in the third draw}
\displaystyle D=\text{An ace in the fourth draw}
\displaystyle \text{Now,}
\displaystyle P(A)=\frac{4}{52}=\frac{1}{13}
\displaystyle P(B\mid A)=\frac{3}{51}=\frac{1}{17}
\displaystyle P(C\mid A\cap B)=\frac{2}{50}=\frac{1}{25}
\displaystyle P(D\mid A\cap B\cap C)=\frac{1}{49}
\displaystyle \therefore\ \text{Required probability}=P(A\cap B\cap C\cap D)=P(A)\times P(B\mid A)\times P(C\mid A\cap B)\times P(D\mid A\cap B\cap C)
\displaystyle =\frac{1}{13}\times\frac{1}{17}\times\frac{1}{25}\times\frac{1}{49}
\displaystyle =\frac{1}{270725}
\displaystyle \text{In case of kings, the required probability will be }=\frac{1}{270725}

\displaystyle \textbf{Question 3: }~\text{Find the chance of drawing }2\text{ white balls in succession from a bag} \\ \text{containing }5\text{ red and }7\text{ white balls, the ball first drawn not being replaced.}
\displaystyle \text{Answer:}
\displaystyle \text{Consider the given events.}
\displaystyle A=\text{A white ball in the first draw}
\displaystyle B=\text{A white ball in the second draw}
\displaystyle \text{Now,}
\displaystyle P(A)=\frac{7}{12}
\displaystyle P(B\mid A)=\frac{6}{11}
\displaystyle \therefore\ \text{Required probability}=P(A\cap B)=P(A)\times P(B\mid A)=\frac{7}{12}\times\frac{6}{11}=\frac{7}{22}

\displaystyle \textbf{Question 4: }~\text{A bag contains }25\text{ tickets, numbered from }1\text{ to }25.\text{ A ticket is drawn} \\ \text{and then another ticket is drawn without replacement. Find the probability that both} \\ \text{tickets will show even numbers.}
\displaystyle \text{Answer:}
\displaystyle \text{There are }12\text{ even numbers between }1\text{ to }25.
\displaystyle \text{Consider the given events.}
\displaystyle A=\text{An even number ticket in the first draw}
\displaystyle B=\text{An even number ticket in the second draw}
\displaystyle \text{Now,}
\displaystyle P(A)=\frac{12}{25}
\displaystyle P(B\mid A)=\frac{11}{24}
\displaystyle \therefore\ \text{Required probability}=P(A\cap B)=P(A)\times P(B\mid A)=\frac{12}{25}\times\frac{11}{24}=\frac{11}{50}

\displaystyle \textbf{Question 5: }~\text{From a deck of cards, three cards are drawn one by one without} \\ \text{replacement. Find the probability that each time it is a card of spade.}
\displaystyle \text{Answer:}
\displaystyle \text{Consider the events}
\displaystyle A=\text{An ace in the first draw}
\displaystyle B=\text{An ace in the second draw}
\displaystyle C=\text{Getting an ace in the third draw}
\displaystyle \text{Now,}
\displaystyle P(A)=\frac{13}{52}=\frac{1}{4}
\displaystyle P(B\mid A)=\frac{12}{51}=\frac{4}{17}
\displaystyle P(C\mid A\cap B)=\frac{11}{50}
\displaystyle \therefore\ \text{Required probability}=P(A\cap B\cap C)
\displaystyle =P(A)\times P(B\mid A)\times P(C\mid A\cap B)
\displaystyle =\frac{1}{4}\times\frac{4}{17}\times\frac{11}{50}
\displaystyle =\frac{11}{850}

\displaystyle \textbf{Question 6: }~\text{Two cards are drawn without replacement from a pack of }52\text{ cards. Find} \\ \text{the probability that (i) both are kings (ii) the first is a king and the second is an ace} \\ \text{(iii) the first is a heart and second is red.}
\displaystyle \text{Answer:}
\displaystyle \text{(i)  }
\displaystyle \text{Consider the given events}
\displaystyle A=\text{A king in the first draw}
\displaystyle B=\text{A king in the second draw}
\displaystyle \text{Now,}
\displaystyle P(A)=\frac{4}{52}=\frac{1}{13}
\displaystyle P(B\mid A)=\frac{3}{51}=\frac{1}{17}
\displaystyle \therefore\ \text{Required probability}=P(A\cap B)
\displaystyle =P(A)\times P(B\mid A)
\displaystyle =\frac{1}{13}\times\frac{1}{17}
\displaystyle =\frac{1}{221}
\displaystyle \text{(ii)  }
\displaystyle \text{Consider the given events}
\displaystyle A=\text{A king in the first draw}
\displaystyle B=\text{An ace in the second draw}
\displaystyle \text{Now,}
\displaystyle P(A)=\frac{4}{52}=\frac{1}{13}
\displaystyle P(B\mid A)=\frac{4}{51}
\displaystyle \therefore\ \text{Required probability}=P(A\cap B)
\displaystyle =P(A)\times P(B\mid A)
\displaystyle =\frac{1}{13}\times\frac{4}{51}
\displaystyle =\frac{4}{663}
\displaystyle \text{(iii)  }
\displaystyle \text{Consider the given events.}
\displaystyle A=\text{A heart in the first throw}
\displaystyle B=\text{A red card in the second throw}
\displaystyle \text{Now,}
\displaystyle P(A)=\frac{13}{52}=\frac{1}{4}
\displaystyle P(B\mid A)=\frac{25}{51}
\displaystyle \therefore\ \text{Required probability}=P(A\cap B)
\displaystyle =P(A)\times P(B\mid A)
\displaystyle =\frac{1}{4}\times\frac{25}{51}
\displaystyle =\frac{25}{204}

\displaystyle \textbf{Question 7: }~\text{A bag contains }20\text{ tickets, numbered from }1\text{ to }20.\text{ Two tickets are drawn} \\ \text{without replacement. What is the probability that the first ticket has an even number} \\ \text{and the second an odd number.}
\displaystyle \text{Answer:}
\displaystyle \text{There are }10\text{ even numbers and }10\text{ odd numbers between }1\text{ to }20.
\displaystyle \text{Consider the given events.}
\displaystyle A=\text{An even number in the first draw}
\displaystyle B=\text{An odd number in the second draw}
\displaystyle \text{Now,}
\displaystyle P(A)=\frac{10}{20}=\frac{1}{2}
\displaystyle P(B\mid A)=\frac{10}{19}
\displaystyle \therefore\ \text{Required probability}=P(A\cap B)=P(A)\times P(B\mid A)=\frac{1}{2}\times\frac{10}{19}=\frac{5}{19}

\displaystyle \textbf{Question 8: }~\text{An urn contains }3\text{ white, }4\text{ red and }5\text{ black balls. Two balls are drawn} \\ \text{one by one without replacement. What is the probability that at least one ball is black?}
\displaystyle \text{Answer:}
\displaystyle \text{Consider the given events.}
\displaystyle A=\text{A white or red ball in the first draw}
\displaystyle B=\text{A white or red ball in the second draw}
\displaystyle \text{Now,}
\displaystyle P(A)=\frac{7}{12}
\displaystyle P(B\mid A)=\frac{6}{11}
\displaystyle \therefore P(A\cap B)=P(A)\times P(B\mid A)
\displaystyle =\frac{7}{12}\times\frac{6}{11}
\displaystyle =\frac{7}{22}
\displaystyle \therefore\ \text{Required probability}=1-P(A\cap B)
\displaystyle =1-\frac{7}{22}
\displaystyle =\frac{15}{22}

\displaystyle \textbf{Question 9: }~\text{A bag contains }5\text{ white, }7\text{ red and }3\text{ black balls. If three balls} \\ \text{are drawn one by one without replacement, find the probability that none is red.}
\displaystyle \text{Answer:}
\displaystyle \text{Consider the given events.}
\displaystyle A=\text{A white or black ball in the first draw}
\displaystyle B=\text{A white or black ball in the second draw}
\displaystyle C=\text{A white or black ball in the third draw}
\displaystyle \text{Now,}
\displaystyle P(A)=\frac{8}{15}
\displaystyle P(B\mid A)=\frac{7}{14}=\frac{1}{2}
\displaystyle P(C\mid A\cap B)=\frac{6}{13}
\displaystyle \therefore\ \text{Required probability}=P(A\cap B\cap C)
\displaystyle =P(A)\times P(B\mid A)\times P(C\mid A\cap B)
\displaystyle =\frac{8}{15}\times\frac{1}{2}\times\frac{6}{13}
\displaystyle =\frac{8}{65}

\displaystyle \textbf{Question 10: }~\text{A card is drawn from a well-shuffled deck of }52\text{ cards and then a second} \\ \text{card is drawn. Find the probability that the first card is a heart and the second card is a} \\ \text{diamond if the first card is not replaced.}
\displaystyle \text{Answer:}
\displaystyle \text{Consider the given events.}
\displaystyle A=\text{A heart in the first draw}
\displaystyle B=\text{A diamond in the second draw}
\displaystyle \text{Now,}
\displaystyle P(A)=\frac{13}{52}=\frac{1}{4}
\displaystyle P(B\mid A)=\frac{13}{51}
\displaystyle \therefore\ \text{Required probability}=P(A\cap B)=P(A)\times P(B\mid A)=\frac{1}{4}\times\frac{13}{51}=\frac{13}{204}

\displaystyle \textbf{Question 11: }~\text{An urn contains }10\text{ black and }5\text{ white balls. Two balls are drawn} \\ \text{from the urn one after the other without replacement. What is the probability that} \\ \text{both drawn balls are black? }
\displaystyle \text{Answer:}
\displaystyle \text{Consider the given events.}
\displaystyle A=\text{A black ball in the first draw}
\displaystyle B=\text{A black ball in the second draw}
\displaystyle \text{Now,}
\displaystyle P(A)=\frac{10}{15}=\frac{2}{3}
\displaystyle P(B\mid A)=\frac{9}{14}
\displaystyle \therefore\ \text{Required probability}=P(A\cap B)=P(A)\times P(B\mid A)=\frac{2}{3}\times\frac{9}{14}=\frac{3}{7}

\displaystyle \textbf{Question 12: }~\text{Three cards are drawn successively, without replacement from a pack} \\ \text{of 52 well shuffled cards. What is the probability that first two cards are kings and third} \\ \text{card drawn is an ace? }
\displaystyle \text{Answer:}
\displaystyle \text{Consider the given events.}
\displaystyle A=\text{A king in the first draw}
\displaystyle B=\text{A king in the second draw}
\displaystyle C=\text{An ace in the third draw}
\displaystyle \text{Now,}
\displaystyle P(A)=\frac{4}{52}=\frac{1}{13}
\displaystyle P(B\mid A)=\frac{3}{51}=\frac{1}{17}
\displaystyle P(C\mid A\cap B)=\frac{4}{50}=\frac{2}{25}
\displaystyle \therefore\ \text{Required probability}=P(A\cap B\cap C)
\displaystyle =P(A)\times P(B\mid A)\times P(C\mid A\cap B)
\displaystyle =\frac{1}{13}\times\frac{1}{17}\times\frac{2}{25}
\displaystyle =\frac{2}{5525}

\displaystyle \textbf{Question 13: }~\text{A box of oranges is inspected by examining three randomly selected oranges} \\ \text{drawn without replacement. If all the three oranges are good, the box is approved for sale} \\ \text{otherwise it is rejected. Find the probability that a box containing }15\text{ oranges out of which } \\ 12\text{ are good and } 3\text{ are bad ones will} \text{be approved for sale.}
\displaystyle \text{Answer:}
\displaystyle \text{Consider the given events.}
\displaystyle A=\text{A good orange in the first draw}
\displaystyle B=\text{A good orange in the second draw}
\displaystyle C=\text{A good orange in the third draw}
\displaystyle \text{Now,}
\displaystyle P(A)=\frac{12}{15}=\frac{4}{5}
\displaystyle P(B\mid A)=\frac{11}{14}
\displaystyle P(C\mid A\cap B)=\frac{10}{13}
\displaystyle \therefore\ \text{Required probability}=P(A\cap B\cap C)
\displaystyle =P(A)\times P(B\mid A)\times P(C\mid A\cap B)
\displaystyle =\frac{4}{5}\times\frac{11}{14}\times\frac{10}{13}
\displaystyle =\frac{44}{91}

\displaystyle \textbf{Question 14: }~\text{A bag contains }4\text{ white, }7\text{ black and }5\text{ red balls. Three balls} \\ \text{are drawn one after the other without replacement. Find the probability that the} \\ \text{balls drawn are white, black and red respectively.}
\displaystyle \text{Answer:}
\displaystyle \text{Consider the given events.}
\displaystyle A=\text{A white ball in the first draw}
\displaystyle B=\text{A black ball in the second draw}
\displaystyle C=\text{A red ball in the third draw}
\displaystyle \text{Now,}
\displaystyle P(A)=\frac{4}{16}=\frac{1}{4}
\displaystyle P(B\mid A)=\frac{7}{15}
\displaystyle P(C\mid A\cap B)=\frac{5}{14}
\displaystyle \therefore\ \text{Required probability}=P(A\cap B\cap C)=P(A)\times P(B\mid A)\times P(C\mid A\cap B)
\displaystyle =\frac{1}{4}\times\frac{7}{15}\times\frac{5}{14}
\displaystyle =\frac{1}{24}

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