Boxes B_1, B_2,and B_3 contain balls as given | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $W_i$ and $B_i$ denote the events of drawing a white ball and a black ball from box $B_i$ respectively. The probabilities are:For $B_1$: $P(W_

Vedclass · Accepted Answer

$\frac{5}{12}$

Vedclass · Answer

(A) Let $W_i$ and $B_i$ denote the events of drawing a white ball and a black ball from box $B_i$ respectively. The probabilities are:For $B_1$: $P(W_1) = \frac{1}{3}, P(B_1) = \frac{2}{3}$For $B_2$: $P(W_2) = \frac{3}{4}, P(B_2) = \frac{1}{4}$For $B_3$: $P(W_3) = \frac{2}{5}, P(B_3) = \frac{3}{5}$We want the probability of drawing two black balls and one white ball. This can happen in three mutually exclusive ways:$1$. $(W_1, B_2, B_3)$: $P(W_1) 	imes P(B_2) 	imes P(B_3) = \frac{1}{3} 	imes \frac{1}{4} 	imes \frac{3}{5} = \frac{3}{60}$$2$. $(B_1, W_2, B_3)$: $P(B_1) 	imes P(W_2) 	imes P(B_3) = \frac{2}{3} 	imes \frac{3}{4} 	imes \frac{3}{5} = \frac{18}{60}$$3$. $(B_1, B_2, W_3)$: $P(B_1) 	imes P(B_2) 	imes P(W_3) = \frac{2}{3} 	imes \frac{1}{4} 	imes \frac{2}{5} = \frac{4}{60}$The total probability is the sum of these probabilities:$P = \frac{3}{60} + \frac{18}{60} + \frac{4}{60} = \frac{25}{60} = \frac{5}{12}$

Boxes $B_1, B_2$,and $B_3$ contain balls as given below:
Box White Black
$B_1$ $1$ $2$
$B_2$ $3$ $1$
$B_3$ $2$ $3$

One ball is drawn at random from each box. Then,among the balls drawn,the probability that two are black and one is white,is

Similar Questions

The probability that a non-leap year contains $53$ Sundays is

Three coins are tossed once. Let $A$ denote the event 'three heads show',$B$ denote the event 'two heads and one tail show',$C$ denote the event 'three tails show',and $D$ denote the event 'a head shows on the first coin'. Which events are mutually exclusive?

An integer is chosen at random and squared. The probability that the last digit of the square is $1$ or $5$ is

If $X_1, X_2, \ldots, X_n$ are $n$ independent events such that $P(X_r) = \frac{1}{r+1}$ for $r = 1, 2, \ldots, n$,then the probability that none of the $n$ events occur is

Let $S$ be the sample space of the random experiment of throwing simultaneously two unbiased dice and $E_k = \{(a, b) \in S : ab = k\}$. If $p_k = P(E_k)$, then which of the following is correct?

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Boxes $B_1, B_2$,and $B_3$ contain balls as given below:BoxWhiteBlack$B_1$$1$$2$$B_2$$3$$1$$B_3$$2$$3$One ball is drawn at random from each box. Then,among the balls drawn,the probability that two are black and one is white,is