For k=1,2,3 the box B_ k contains k red balls and (k+1) white balls. Let P (B_ 1 )=(1)/(2), P (B_ 2 )=(1)/(3) and P(B_ 3 )=(1)/(6). A box is selected at random and a ball is drawn from it. If a red ball is drawn, then

Correct answer: B. (14)/(39) In a box, B_ 1 =1 R, 2 W ; B_ 2 =2 R, 3 W and B_ 3 =3 R, 4 W Also, given that, P(B_ 1 )=(1)/(2), P(B_ 2 )=(1)/(3) aligned & and P(B_ 3 )=(1)/(6) & P( B_ 2 R )= P(B_ 2 ) P( R B_ 2 ) P(B_ 1 ) P( R B_ 1 )+P(B_ 2 ) P( R B_ 2 )+P(B_ 3 ) P( R B_ 3 ) =& ((1)/(3) × (2)/(5))/((1)/(2) × (1)/(3)+(1)/(3) × (2)/(5)+(1)/(6) × (3)/(7))=((2)/(15))/((1)/(6)+(2)/(15)+(1)/(14))=(14)/(39) . aligned

BITSAT 2013

Exam: BITSAT
Year: 2013
Subject: Mathematics
Chapter: Probability

BITSAT 2013 · Mathematics · Probability

For k=1,2,3 the box B_ k contains k red balls and (k+1) white...

For

k=1,2,3

the box

\mathrm{B}_{k}

contains

k

red balls and

(k+1)

white balls. Let

\mathrm{P}

\left(\mathrm{B}_{1}\right)=\frac{1}{2}, \mathrm{P}

\left(\mathrm{B}_{2}\right)=\frac{1}{3}

and

\mathrm{P}\left(\mathrm{B}_{3}\right)=\frac{1}{6}. \mathrm{A}

box is selected at random and a ball is drawn from it. If a red ball is drawn, then the probability that it has come from box

\mathrm{B}_{2},

Options

A $\frac{35}{78}$
B $\frac{14}{39}$
C $\frac{10}{13}$
D $\frac{12}{13}$

Correct answer

B. $\frac{14}{39}$

Step-by-step solution

\operatorname{In}

a box,

B_{1}=1 R, 2 W ; B_{2}=2 R, 3 W and B_{3}=3 R, 4 W

Also, given that,

\mathrm{P}\left(\mathrm{B}_{1}\right)=\frac{1}{2}, \mathrm{P}\left(\mathrm{B}_{2}\right)=\frac{1}{3}

\begin{aligned}& \text{\text{ and } }P\left(B_{3}\right)=\frac{1}{6}\\ \therefore \quad & P\left(\frac{B_{2}}{R}\right)=\frac{P\left(B_{2}\right) P\left(\frac{R}{B_{2}}\right)}{P\left(B_{1}\right) P\left(\frac{R}{B_{1}}\right)+P\left(B_{2}\right) P\left(\frac{R}{B_{2}}\right)+P\left(B_{3}\right) P\left(\frac{R}{B_{3}}\right)}\\ =& \frac{\frac{1}{3} \times \frac{2}{5}}{\frac{1}{2} \times \frac{1}{3}+\frac{1}{3} \times \frac{2}{5}+\frac{1}{6} \times \frac{3}{7}}=\frac{\frac{2}{15}}{\frac{1}{6}+\frac{2}{15}+\frac{1}{14}}=\frac{14}{39} . \end{aligned}

Practice Probability on Quantrex Academy →

More from Probability

The probability of getting 10 in a single throw of three fair dice is BITSAT 2024 In a binomial distribution, the mean is 4 and variance is 3. Then, its mode is BITSAT 2024 The probability that certain electronic component fails when first used is 0.10. If it… BITSAT 2024 Given below is the distribution of a random variable X If α=P(X 2) , then α: β= BITSAT 2024 A book contains 1000 pages. A page is chosen at random. The probability that the sum of… BITSAT 2024 For two events A and B , if P(A)=P( A B )=(1)/(4) and P((B)/(A))=(1)/(2) , then which of… BITSAT 2024 If A and B are mutually exclusive events and if P(B)=(1)/(3), P(A B)=(13)/(21) , then… BITSAT 2023 A coin is tossed twice. Then, the probability that atleast one tail occurs is BITSAT 2023 If E and F are events such that 0 < P(F) < 1 , then BITSAT 2023 The probability of getting sum more than 7 when a pair of dice are thrown is: BITSAT 2023 The probability that a card drawn from a pack of 52 cards will be a diamond or king is: BITSAT 2023 Bag P contains 6 red and 4 blue balls and bag Q contains 5 red and 6 blue balls. A ball… BITSAT 2023 If P(B)=(3)/(5), P(A B)=(1)/(2) and P(A B)=(4)/(5) , then P(A B)^ +P(A^ B)= BITSAT 2023 A six faced die is a biased one. It is thrice more likely to show an odd numbers than… BITSAT 2022 Girl students constitute 10 \% of I year and 5 \% of II year at Roorkee University.… BITSAT 2021 The mean and variance of a random variable X having binomial distribution are 4 and 2… BITSAT 2021 Full Probability list Probability JEE Maths PYQs JEE Mathematics PYQs All BITSAT PYQs