Question

Two chips are drawn at random and without replacement from a bag containing two blue chips and two red chips. Events A and B are defined as follows. A: {At least one of the chips is blue}. B: {Both chips are red}. The events A and B are mutually exclusive.

Answer 1

Answer:

Yes, The events A and B are mutually exclusive event.

Step-by-step explanation:

Two chips are drawn at random and without replacement from a bag containing two blue chips and two red chips.

Total chips (n ) = 4

Two chips are chooses from 4 chips then the total possibilities are = $\binom{4}{2}$

total outcomes = $\frac{4!}{(2!)(2!)}$ = 6

Let $b_{1}$ and $b_{2}$ denote blue chips and $r_{1}$ and $r_{2}$ denote red chips.

A: {At least one of the chips is blue} = { $b_{1}$$b_{2$ , $b_{1}r_{1}$ , $b_{1}r_{2}$ , $b_{2}r_{1}$, $b_{2}r_{2}$ }

P( A) = $\frac{5}{6}$

B: {Both chips are red} = { $r_{1}r_{2}$ }

P( B)  = $\frac{1}{6}$

$(A\bigcup B)$  = { $b_{2}b_{1}$ , $b_{1}r_{1} , b_{1}r_{2} , b_{2}r_{1} , b_{2}r_{2} ,r_{1}r_{2}$ }

$P(A\bigcup B)$ = $\frac{6}{6}$

If A and B are mutuall exclusive

$P(A\bigcup B)$ = P( A) + P( B)

P( A) + P( B)  = $\frac{5}{6} + \frac{1}{6}$ = 1

$P(A\bigcup B)$ = $\frac{6}{6}$ = 1

Hence

$P(A\bigcup B)$ = P( A) + P( B)

Then A and B are mutuall exclusive