Question

At a recent graduation at a naval flight school, 18 Marines, 10 members of the Navy, and 3 members of the Coast Guard got their wings. Choose 3 pilots at random to feature on a training brochure. Find the probability that there will be:

a.) 1 of each.
b.) 0 members of the Navy.
c.) 3 Marines.

How would I solve parts A,B, and C ?

Answer 1

Explanation:

Given:

n(M) = 18

n(N) = 10

n(C) = 3

n(S) = 18 + 10 + 3 = 31

a)

A = {choose 3 pilots at random}

`n(A)=\ _(31)C_3`

`=(31!)/(3!(31-3)!)`

`=4495`

B = {1 of each}

`n(B)=\ _(18)C_1*_(10)C_1*_3C_1`

`=18*10*3`

`=540`

`P(B)=(n(B))/(n(A))`

`=(540)/(4495)`

`=(108)/(899)`

b)

C = {1 Navy}

D = {2 Navy}

E = {3 Navy}

F = {0 Navy} → n(F) = n(A) - (n(C) + n(D) + n(E))

`n(C)=\ _(10)C_1* _((31-10))C_2`

`=10*(21!)/(2!(31-2)!)`

`=2100`

`n(D)=\ _(10)C_2* _((31-10))C_1`

`=(10!)/(2!(10-2)!)*21`

`=945`

`n(E)=\ _(10)C_3`

`=(10!)/(3!(10-3)!)`

`=120`

`n(F)=4495-(2100+945+120)`

`=1330`

`P(F)=(n(F))/(n(A))`

`=(1330)/(4495)`

`=(266)/(899)`

c)

G = {3 Marine}

`n(G)=\ _(18)C_3`

`=(18!)/(3!(18-3)!)`

`=816`

`P(G)=(n(G))/(n(A))`

`=(816)/(4495)`