A box contains 1 red and 3 identical white ba | vedclass

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

Question

(B) The box contains $1$ red ball $(R)$ and $3$ identical white balls $(W)$.Since the balls are drawn without replacement,the first ball can be red or

Vedclass · Accepted Answer

$S = \{ RW, WR, WW \}$

Vedclass · Answer

(B) The box contains $1$ red ball $(R)$ and $3$ identical white balls $(W)$.Since the balls are drawn without replacement,the first ball can be red or white.If the first ball is red,the second must be white (as there is only $1$ red ball).If the first ball is white,the second can be red or white.Thus,the possible outcomes are:$1$. First is red,second is white: $(R, W)$$2$. First is white,second is red: $(W, R)$$3$. First is white,second is white: $(W, W)$Therefore,the sample space is $S = \{ RW, WR, WW \}$.

$A$ box contains $1$ red and $3$ identical white balls. Two balls are drawn at random in succession without replacement. Write the sample space for this experiment.

Similar Questions

The probability that at least one of $A$ and $B$ occurs is $0.6$. If $A$ and $B$ occur simultaneously with probability $0.3$,then $P(A') + P(B') = $

The probability of getting a sum of $3$,$5$,or $11$ when throwing two dice is:

If a dice is thrown twice,then the probability of getting $1$ in the first throw only is

In a single throw of two dice,what is the probability of obtaining an odd number as the sum of the numbers on the two dice?

If $A$,$B$,and $C$ are mutually exclusive and exhaustive events of a sample space $S$ such that $P(B) = \frac{3}{2} P(A)$ and $P(C) = \frac{1}{2} P(B)$,then $P(A) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module