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

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

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 =
`(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) =
`(5)/(6)`

B: {Both chips are red} = {
`r_(1)r_(2)` }

P( B) =
`(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)` =
`(6)/(6)`

If A and B are mutuall exclusive

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

P( A) + P( B) =
`(5)/(6) + (1)/(6)` = 1

`P(A\bigcup B)` =
`(6)/(6)` = 1

Hence

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

Then A and B are mutuall exclusive