Question

A box contains 3 green marbles, 5 blue marbles, and 7 red marbles. Three marbles are selected at random from the box, one at a time, without replacement. Find the probability that the first two marbles selected are not red, and the last marble is red. Round your answer to four decimal places.A box contains 3 green marbles, 5 blue marbles, and 7 red marbles. Three marbles are selected at random from the box, one at a time, without replacement. Find the probability that the first two marbles selected are not red, and the last marble is red. Round your answer to four decimal places.

Answer 1

The probability is
`P(K)  =   ( 28 )/(195)`

Explanation:

From the question we are told that

The number of green marbles is
`n_g  =  3`

The number of red marbles is
`n_b  =  5`

The number of red marbles is
`n_r  =  7`

Generally the total number of marbles is mathematically represented as

`n_t  =  n_r  +  n_g + n_ b`

`n_t  =  7 +  3 + 5`

`n_t  = 5`

Generally total number of marbles that are not red is

`n_k  =  n_g +  n_ b`

=>
`n_k  =   3 +   5`

=>
`n_k  =  8`

The probability of the first ball not being red is mathematically represented as

`P(r') =  (n_k)/(n_t)`

=>
`P(r') =  ( 8)/(15)`

The probability of the second ball not being red is mathematically represented as

`P(r'') =  (n_k - 1)/(n_t -1)`

=>
`P(r'') =  ( 8 -1 )/(15-1)` (the subtraction is because the marbles where selected without replacement )

=>
`P(r'') =  ( 7 )/(14)`

The probability that the first two balls is not red is mathematically represented as

`P(R) =  P(r') *  P(r'')`

=>
`P(R) =  ( 8)/(15) *   ( 7 )/(14)`

=>
`P(R) =  ( 8 )/(30)`

The probability of the third ball being red is mathematically represented as

`P(r) =  (n_r)/( n_t -2)` (the subtraction is because the marbles where selected without replacement )

`P(r) =  (7)/( 15 -2)`

=>
`P(r) =  (7)/( 13)`

Generally the probability of the first two marble not being red and the third marble being red is mathematically represented as

`P(K)  =  P(R) * P(r)`

`P(K)  =   ( 8 )/(30) * (7)/( 13)`

=>
`P(K)  =   ( 28 )/(195)`