A box contains 3 white and 2 red balls. A bal | vedclass

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

Question

(B) The total number of balls is $3 + 2 = 5$.Let $W_1$ be the event that the first ball is white and $R_1$ be the event that the first ball is red.Let

Vedclass · Accepted Answer

$\frac{2}{5}$

Vedclass · Answer

(B) The total number of balls is $3 + 2 = 5$.Let $W_1$ be the event that the first ball is white and $R_1$ be the event that the first ball is red.Let $R_2$ be the event that the second ball is red.The second ball can be red in two mutually exclusive ways:$(i)$ First ball is white and second ball is red: $P(W_1 \cap R_2) = P(W_1) 	imes P(R_2|W_1) = \frac{3}{5} 	imes \frac{2}{4} = \frac{6}{20}$.$(ii)$ First ball is red and second ball is red: $P(R_1 \cap R_2) = P(R_1) 	imes P(R_2|R_1) = \frac{2}{5} 	imes \frac{1}{4} = \frac{2}{20}$.The total probability of the second ball being red is $P(R_2) = P(W_1 \cap R_2) + P(R_1 \cap R_2) = \frac{6}{20} + \frac{2}{20} = \frac{8}{20} = \frac{2}{5}$.

$A$ box contains $3$ white and $2$ red balls. $A$ ball is drawn and another ball is drawn without replacing the first ball. Then,the probability of the second ball being red is:

Similar Questions

From $80$ cards numbered $1$ to $80$, two cards are selected randomly. The probability that both the cards have numbers divisible by $4$ is given by

Find the probability that a two-digit number formed by the digits $1, 2, 3, 4, 5$ is divisible by $4$ (repetition of digits is allowed).

$A$ three-digit number is formed by using the digits $1, 2, 3,$ and $4$. The probability that the number is divisible by $3$ is:

$A$ bag contains $3$ white and $5$ black balls. If one ball is drawn, then the probability that it is black, is

$A$ basket contains $3$ blue and $4$ red balls. If $3$ balls are drawn at random from the basket,then what is the probability that all the $3$ are either blue or red?

Vedclass Test Series

Exam Paper Generator

Online Exam Module