Question 1:
A bag A contains \displaystyle 5 white and \displaystyle 6 black balls. Another bag B contains \displaystyle 4 white and \displaystyle 3 black balls. A ball is transferred from bag A to the bag B and then a ball is taken out of the second bag. Find the probability of this ball being black.
\displaystyle \text{Answer:}
\displaystyle \text{A black ball can be drawn in two mutually exclusive ways:}
\displaystyle \text{(I) By transferring a white ball from bag A to bag B, then drawing a black ball}
\displaystyle \text{(II) By transferring a black ball from bag A to bag B, then drawing a black ball}
\displaystyle \text{Let } E_1, E_2 \text{ and } A \text{ be the events as defined below:}
\displaystyle E_1 = \text{A white ball is transferred from bag A to bag B}
\displaystyle E_2 = \text{A black ball is transferred from bag A to bag B}
\displaystyle A = \text{A black ball is drawn}
\displaystyle \therefore P(E_1)=\frac{5}{11}
\displaystyle P(E_2)=\frac{6}{11}
\displaystyle \text{Now,}
\displaystyle P(A\mid E_1)=\frac{3}{8}
\displaystyle P(A\mid E_2)=\frac{4}{8}
\displaystyle \text{Using the law of total probability, we get}
\displaystyle \text{Required probability}=P(A)=P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)
\displaystyle =\frac{5}{11}\times\frac{3}{8}+\frac{6}{11}\times\frac{4}{8}
\displaystyle =\frac{15}{88}+\frac{24}{88}
\displaystyle =\frac{39}{88}

Question 2:
A purse contains \displaystyle 2 silver and \displaystyle 4 copper coins. A second purse contains \displaystyle 4 silver and \displaystyle 3 copper coins. If a coin is pulled at random from one of the two purses, what is the probability that it is a silver coin?
\displaystyle \text{Answer:}
\displaystyle \text{A silver coin can be drawn in two mutually exclusive ways:}
\displaystyle \text{(I) Selecting purse I and then drawing a silver coin from it}
\displaystyle \text{(II) Selecting purse II and then drawing a silver coin from it}
\displaystyle \text{Let } E_1, E_2 \text{ and } A \text{ be the events as defined below:}
\displaystyle E_1=\text{Selecting purse I}
\displaystyle E_2=\text{Selecting purse II}
\displaystyle A=\text{Drawing a silver coin}
\displaystyle \text{It is given that one of the purses is selected randomly}
\displaystyle \therefore P(E_1)=\frac{1}{2}
\displaystyle P(E_2)=\frac{1}{2}
\displaystyle \text{Now,}
\displaystyle P(A\mid E_1)=\frac{2}{6}=\frac{1}{3}
\displaystyle P(A\mid E_2)=\frac{4}{7}
\displaystyle \text{Using the law of total probability, we get}
\displaystyle \text{Required probability}=P(A)=P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)
\displaystyle =\frac{1}{2}\times\frac{1}{3}+\frac{1}{2}\times\frac{4}{7}
\displaystyle =\frac{1}{6}+\frac{2}{7}
\displaystyle =\frac{7+12}{42}=\frac{19}{42}

Question 3:
One bag contains \displaystyle 4 yellow and \displaystyle 5 red balls. Another bag contains \displaystyle 6 yellow and \displaystyle 3 red balls. A ball is transferred from the first bag to the second bag and then a ball is drawn from the second bag. Find the probability that ball drawn is yellow.
\displaystyle \text{Answer:}
\displaystyle \text{A yellow ball can be drawn in two mutually exclusive ways:}
\displaystyle \text{(I) By transferring a red ball from first to second bag, then drawing a yellow ball}
\displaystyle \text{(II) By transferring a yellow ball from first to second bag, then drawing a yellow ball}
\displaystyle \text{Let } E_1, E_2 \text{ and } A \text{ be the events as defined below:}
\displaystyle E_1=\text{A red ball is transferred from first to second bag}
\displaystyle E_2=\text{A yellow ball is transferred from first to second bag}
\displaystyle A=\text{A yellow ball is drawn}
\displaystyle \therefore P(E_1)=\frac{5}{9}
\displaystyle P(E_2)=\frac{4}{9}
\displaystyle \text{Now,}
\displaystyle P(A\mid E_1)=\frac{6}{10}
\displaystyle P(A\mid E_2)=\frac{7}{10}
\displaystyle \text{Using the law of total probability, we get}
\displaystyle \text{Required probability}=P(A)=P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)
\displaystyle =\frac{5}{9}\times\frac{6}{10}+\frac{4}{9}\times\frac{7}{10}
\displaystyle =\frac{30}{90}+\frac{28}{90}
\displaystyle =\frac{58}{90}=\frac{29}{45}

Question 4:
A bag contains \displaystyle 3 white and \displaystyle 2 black balls and another bag contains \displaystyle 2 white and \displaystyle 4 black balls. One bag is chosen at random. From the selected bag, one ball is drawn. Find the probability that the ball drawn is white.
\displaystyle \text{Answer:}
\displaystyle \text{A white ball can be drawn in two mutually exclusive ways:}
\displaystyle \text{(I) Selecting bag I and then drawing a white ball from it}
\displaystyle \text{(II) Selecting bag II and then drawing a white ball from it}
\displaystyle \text{Let } E_1, E_2 \text{ and } A \text{ be the events as defined below:}
\displaystyle E_1=\text{Selecting bag I}
\displaystyle E_2=\text{Selecting bag II}
\displaystyle A=\text{Drawing a white ball}
\displaystyle \text{It is given that one of the bags is selected randomly}
\displaystyle \therefore P(E_1)=\frac{1}{2}
\displaystyle P(E_2)=\frac{1}{2}
\displaystyle \text{Now,}
\displaystyle P(A\mid E_1)=\frac{3}{5}
\displaystyle P(A\mid E_2)=\frac{2}{6}=\frac{1}{3}
\displaystyle \text{Using the law of total probability, we get}
\displaystyle \text{Required probability}=P(A)=P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)
\displaystyle =\frac{1}{2}\times\frac{3}{5}+\frac{1}{2}\times\frac{1}{3}
\displaystyle =\frac{3}{10}+\frac{1}{6}
\displaystyle =\frac{9+5}{30}=\frac{14}{30}=\frac{7}{15}

Question 5:
The contents of three bags I, II and III are as follows:
Bag I : \displaystyle 1 white, \displaystyle 2 black and \displaystyle 3 red balls,
Bag II : \displaystyle 2 white, \displaystyle 1 black and \displaystyle 1 red ball;
Bag III : \displaystyle 4 white, \displaystyle 5 black and \displaystyle 3 red balls.
A bag is chosen at random and two balls are drawn. What is the probability that the balls are white and red?
\displaystyle \text{Answer:}
\displaystyle \text{A white ball and a red ball can be drawn in three mutually exclusive ways:}
\displaystyle \text{(I) Selecting bag I and then drawing a white and a red ball from it}
\displaystyle \text{(II) Selecting bag II and then drawing a white and a red ball from it}
\displaystyle \text{(III) Selecting bag III and then drawing a white and a red ball from it}
\displaystyle \text{Let } E_1,E_2,E_3 \text{ and } A \text{ be the events as defined below:}
\displaystyle E_1=\text{Selecting bag I}
\displaystyle E_2=\text{Selecting bag II}
\displaystyle E_3=\text{Selecting bag III}
\displaystyle A=\text{Drawing a white and a red ball}
\displaystyle \text{It is given that one of the bags is selected randomly}
\displaystyle \therefore P(E_1)=\frac{1}{3}
\displaystyle P(E_2)=\frac{1}{3}
\displaystyle P(E_3)=\frac{1}{3}
\displaystyle \text{Now,}
\displaystyle P(A\mid E_1)=\frac{\binom{1}{1}\binom{3}{1}}{\binom{6}{2}}=\frac{3}{15}
\displaystyle P(A\mid E_2)=\frac{\binom{2}{1}\binom{1}{1}}{\binom{4}{2}}=\frac{2}{6}
\displaystyle P(A\mid E_3)=\frac{\binom{4}{1}\binom{3}{1}}{\binom{12}{2}}=\frac{12}{66}
\displaystyle \text{Using the law of total probability, we get}
\displaystyle \text{Required probability}=P(A)=P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)+P(E_3)P(A\mid E_3)
\displaystyle =\frac{1}{3}\times\frac{3}{15}+\frac{1}{3}\times\frac{2}{6}+\frac{1}{3}\times\frac{12}{66}
\displaystyle =\frac{1}{15}+\frac{1}{9}+\frac{2}{33}
\displaystyle =\frac{33+55+30}{495}=\frac{118}{495}

Question 6:
An unbiased coin is tossed. If the result is a head, a pair of unbiased dice is rolled and the sum of the numbers obtained is noted. If the result is a tail, a card from a well shuffled pack of eleven cards numbered \displaystyle 2,3,4,\ldots,12 is picked and the number on the card is noted. What is the probability that the noted number is either \displaystyle 7 or \displaystyle 8?
\displaystyle \text{Answer:}
\displaystyle \text{Let } E_1,E_2 \text{ and } A \text{ be the events as defined below:}
\displaystyle E_1=\text{The coin shows a head}
\displaystyle E_2=\text{The coin shows a tail}
\displaystyle A=\text{The noted number is }7\text{ or }8
\displaystyle \therefore P(E_1)=\frac{1}{2}
\displaystyle P(E_2)=\frac{1}{2}
\displaystyle \text{Now,}
\displaystyle P(A\mid E_1)=\frac{11}{36}
\displaystyle P(A\mid E_2)=\frac{2}{11}
\displaystyle \text{Using the law of total probability, we get}
\displaystyle \text{Required probability}=P(A)=P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)
\displaystyle =\frac{1}{2}\times\frac{11}{36}+\frac{1}{2}\times\frac{2}{11}
\displaystyle =\frac{11}{72}+\frac{1}{11}
\displaystyle =\frac{121+72}{792}=\frac{193}{792}

Question 7:
A factory has two machines A and B. Past records show that the machine A produced \displaystyle 60\% of the items of output and machine B produced \displaystyle 40\% of the items. Further \displaystyle 2\% of the items produced by machine A were defective and \displaystyle 1\% produced by machine B were defective. If an item is drawn at random, what is the probability that it is defective?
\displaystyle \text{Answer:}
\displaystyle \text{Let } A,E_1 \text{ and } E_2 \text{ denote the events that the item is defective, machine A is selected and machine B is selected, respectively}
\displaystyle \therefore P(E_1)=\frac{60}{100}
\displaystyle P(E_2)=\frac{40}{100}
\displaystyle \text{Now,}
\displaystyle P(A\mid E_1)=\frac{2}{100}
\displaystyle P(A\mid E_2)=\frac{1}{100}
\displaystyle \text{Using the law of total probability, we get}
\displaystyle \text{Required probability}=P(A)=P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)
\displaystyle =\frac{60}{100}\times\frac{2}{100}+\frac{40}{100}\times\frac{1}{100}
\displaystyle =\frac{120}{10000}+\frac{40}{10000}
\displaystyle =\frac{120+40}{10000}=\frac{160}{10000}=0.016

Question 8:
The bag A contains \displaystyle 8 white and \displaystyle 7 black balls while the bag B contains \displaystyle 5 white and \displaystyle 4 black balls. One ball is randomly picked up from the bag A and mixed up with the balls in bag B. Then a ball is randomly drawn out from it. Find the probability that ball drawn is white.
\displaystyle \text{Answer:}
\displaystyle \text{A white ball can be drawn in two mutually exclusive ways:}
\displaystyle \text{(I) By transferring a black ball from bag A to bag B, then drawing a white ball}
\displaystyle \text{(II) By transferring a white ball from bag A to bag B, then drawing a white ball}
\displaystyle \text{Let } E_1,E_2 \text{ and } A \text{ be events as defined below:}
\displaystyle E_1=\text{A black ball is transferred from bag A to bag B}
\displaystyle E_2=\text{A white ball is transferred from bag A to bag B}
\displaystyle A=\text{A white ball is drawn}
\displaystyle \therefore P(E_1)=\frac{7}{15}
\displaystyle P(E_2)=\frac{8}{15}
\displaystyle \text{Now,}
\displaystyle P(A\mid E_1)=\frac{5}{10}=\frac{1}{2}
\displaystyle P(A\mid E_2)=\frac{6}{10}=\frac{3}{5}
\displaystyle \text{Using the law of total probability, we get}
\displaystyle \text{Required probability}=P(A)=P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)
\displaystyle =\frac{7}{15}\times\frac{1}{2}+\frac{8}{15}\times\frac{3}{5}
\displaystyle =\frac{7}{30}+\frac{8}{25}
\displaystyle =\frac{35+48}{150}=\frac{83}{150}

Question 9:
A bag contains \displaystyle 4 white and \displaystyle 5 black balls and another bag contains \displaystyle 3 white and \displaystyle 4 black balls. A ball is taken out from the first bag and without seeing its colour is put in the second bag. A ball is taken out from the latter. Find the probability that the ball drawn is white.
\displaystyle \text{Answer:}
\displaystyle \text{A white ball can be drawn in two mutually exclusive ways:}
\displaystyle \text{(I) By transferring a black ball from first to second bag, then drawing a white ball}
\displaystyle \text{(II) By transferring a white ball from first to second bag, then drawing a white ball}
\displaystyle \text{Let } E_1,E_2 \text{ and } A \text{ be the events as defined below:}
\displaystyle E_1=\text{A black ball is transferred from first to second bag}
\displaystyle E_2=\text{A white ball is transferred from first to second bag}
\displaystyle A=\text{A white ball is drawn}
\displaystyle \therefore P(E_1)=\frac{5}{9}
\displaystyle P(E_2)=\frac{4}{9}
\displaystyle \text{Now,}
\displaystyle P(A\mid E_1)=\frac{3}{8}
\displaystyle P(A\mid E_2)=\frac{4}{8}=\frac{1}{2}
\displaystyle \text{Using the law of total probability, we get}
\displaystyle \text{Required probability}=P(A)=P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)
\displaystyle =\frac{5}{9}\times\frac{3}{8}+\frac{4}{9}\times\frac{1}{2}
\displaystyle =\frac{15}{72}+\frac{2}{9}
\displaystyle =\frac{15+16}{72}=\frac{31}{72}

Question 10:
One bag contains \displaystyle 4 white and \displaystyle 5 black balls. Another bag contains \displaystyle 6 white and \displaystyle 7 black balls. A ball is transferred from first bag to the second bag and then a ball is drawn from the second bag. Find the probability that the ball drawn is white.
\displaystyle \text{Answer:}
\displaystyle \text{A white ball can be drawn in two mutually exclusive ways:}
\displaystyle \text{(I) By transferring a black ball from first to second bag, then drawing a white ball}
\displaystyle \text{(II) By transferring a white ball from first to second bag, then drawing a white ball}
\displaystyle \text{Let } E_1,E_2 \text{ and } A \text{ be the events as defined below:}
\displaystyle E_1=\text{A black ball is transferred from first to second bag}
\displaystyle E_2=\text{A white ball is transferred from first to second bag}
\displaystyle A=\text{A white ball is drawn}
\displaystyle \therefore P(E_1)=\frac{5}{9}
\displaystyle P(E_2)=\frac{4}{9}
\displaystyle \text{Now,}
\displaystyle P(A\mid E_1)=\frac{6}{14}
\displaystyle P(A\mid E_2)=\frac{7}{14}
\displaystyle \text{Using the law of total probability, we get}
\displaystyle \text{Required probability}=P(A)=P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)
\displaystyle =\frac{5}{9}\times\frac{6}{14}+\frac{4}{9}\times\frac{7}{14}
\displaystyle =\frac{30}{126}+\frac{28}{126}
\displaystyle =\frac{58}{126}=\frac{29}{63}

Question 11:
An urn contains \displaystyle 10 white and \displaystyle 3 black balls. Another urn contains \displaystyle 3 white and \displaystyle 5 black balls. Two are drawn from first urn and put into the second urn and then a ball is drawn from the latter. Find the probability that it is a white ball.
\displaystyle \text{Answer:}
\displaystyle \text{A white ball can be drawn in three mutually exclusive ways:}
\displaystyle \text{(I) By transferring two black balls from first to second urn, then drawing a white ball}
\displaystyle \text{(II) By transferring two white balls from first to second urn, then drawing a white ball}
\displaystyle \text{(III) By transferring a white and a black ball from first to second urn, then drawing a white ball}
\displaystyle \text{Let } E_1,E_2,E_3 \text{ and } A \text{ be the events as defined below:}
\displaystyle E_1=\text{Two black balls are transferred from first to second bag}
\displaystyle E_2=\text{Two white balls are transferred from first to second bag}
\displaystyle E_3=\text{A white and a black ball is transferred from first to second bag}
\displaystyle A=\text{A white ball is drawn}
\displaystyle \therefore P(E_1)=\frac{\binom{3}{2}}{\binom{13}{2}}=\frac{3}{78}
\displaystyle P(E_2)=\frac{\binom{10}{2}}{\binom{13}{2}}=\frac{45}{78}
\displaystyle P(E_3)=\frac{\binom{10}{1}\binom{3}{1}}{\binom{13}{2}}=\frac{30}{78}
\displaystyle \text{Now,}
\displaystyle P(A\mid E_1)=\frac{3}{10}
\displaystyle P(A\mid E_2)=\frac{5}{10}
\displaystyle P(A\mid E_3)=\frac{4}{10}
\displaystyle \text{Using the law of total probability, we get}
\displaystyle \text{Required probability}=P(A)=P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)+P(E_3)P(A\mid E_3)
\displaystyle =\frac{3}{78}\times\frac{3}{10}+\frac{45}{78}\times\frac{5}{10}+\frac{30}{78}\times\frac{4}{10}
\displaystyle =\frac{9}{780}+\frac{225}{780}+\frac{120}{780}
\displaystyle =\frac{354}{780}=\frac{59}{130}

Question 12:
A bag contains \displaystyle 6 red and \displaystyle 8 black balls and another bag contains \displaystyle 8 red and \displaystyle 6 black balls. A ball is drawn from the first bag and without noticing its colour is put in the second bag. A ball is drawn from the second bag. Find the probability that the ball drawn is red in colour.
\displaystyle \text{Answer:}
\displaystyle \text{A red ball can be drawn in two mutually exclusive ways:}
\displaystyle \text{(I) By transferring a black ball from first to second bag, then drawing a red ball}
\displaystyle \text{(II) By transferring a red ball from first to second bag, then drawing a red ball}
\displaystyle \text{Let } E_1,E_2 \text{ and } A \text{ be the events as defined below:}
\displaystyle E_1=\text{A black ball is transferred from first to second bag}
\displaystyle E_2=\text{A red ball is transferred from first to second bag}
\displaystyle A=\text{A red ball is drawn}
\displaystyle \therefore P(E_1)=\frac{8}{14}
\displaystyle P(E_2)=\frac{6}{14}
\displaystyle \text{Now,}
\displaystyle P(A\mid E_1)=\frac{8}{15}
\displaystyle P(A\mid E_2)=\frac{9}{15}
\displaystyle \text{Using the law of total probability, we get}
\displaystyle \text{Required probability}=P(A)=P(E_1)P(A\mid E_1)+P(E_2)P(A\mid E_2)
\displaystyle =\frac{8}{14}\times\frac{8}{15}+\frac{6}{14}\times\frac{9}{15}
\displaystyle =\frac{64}{210}+\frac{54}{210}
\displaystyle =\frac{118}{210}=\frac{59}{105}

Question 13:
Three machines \displaystyle E_1,E_2,E_3 in a certain factory produce \displaystyle 50\%,25\% and \displaystyle 25\% respectively, of the total daily output of electric bulbs. It is known that \displaystyle 4\% of the tubes produced on each of machines \displaystyle E_1 and \displaystyle E_2 are defective, and that \displaystyle 5\% of those produced on \displaystyle E_3 are defective. If one tube is picked up at random from a day’s production, calculate the probability that it is defective.
\displaystyle \text{Answer:}
\displaystyle \text{Let } A \text{ be the event that the tube picked is defective}
\displaystyle \text{We have,}
\displaystyle P(E_1)=50
\displaystyle P(A\mid E_1)=4
\displaystyle \text{Now,}
\displaystyle P(A)=P(E_1)\times P(A\mid E_1)+P(E_2)\times P(A\mid E_2)+P(E_3)\times P(A\mid E_3)
\displaystyle =\frac{1}{2}\times\frac{1}{25}+\frac{1}{4}\times\frac{1}{25}+\frac{1}{4}\times\frac{1}{20}
\displaystyle =\frac{1}{50}+\frac{1}{100}+\frac{1}{80}
\displaystyle =\frac{8+4+5}{400}
\displaystyle =\frac{17}{400}
\displaystyle \text{So, the probability that the picked tube is defective is} \frac{17}{400}

Discover more from ICSE / ISC / CBSE Mathematics Portal for K12 Students

Subscribe to get the latest posts sent to your email.