# Math Help - Discrete Probability Distribution

1. ## Discrete Probability Distribution

from walpole 5.32

From a lot of 10 missiles, 4 are selected at random and fired. If the lot contains 3 defective missiles that will not fire, what is the probability that
a) all 4 will fire
b) at most 2 will not fire ?

I believe this is a hypergeometric distribution.
pls correct me if i am wrong.

a) since all 4 , i took P(x= 0) + .....P(x=4)
4C0 x 6C4 / 10C4 + 4C1 x 6C3 / 10C4 + .....
and i got 13/14. but this does not tally with the answer stated on my worksheet. where did i go wrong ?

b) at most 2 will not fire means P (X<=2) ?

2. Originally Posted by hazel
from walpole 5.32

From a lot of 10 missiles, 4 are selected at random and fired. If the lot contains 3 defective missiles that will not fire, what is the probability that
a) all 4 will fire
b) at most 2 will not fire ?

I believe this is a hypergeometric distribution. Mr F says: Correct.
pls correct me if i am wrong.

a) since all 4 , i took P(x= 0) + .....P(x=4)
4C0 x 6C4 / 10C4 + 4C1 x 6C3 / 10C4 + .....
and i got 13/14. but this does not tally with the answer stated on my worksheet. where did i go wrong ?

b) at most 2 will not fire means P (X<=2) ?
You must define the random variable before doing anything else.

Let X be the random variable 'number of defective missiles in sample'.
X ~ Hypergeometric(N = 10, n = 4, D = 3).

a) Calculate Pr(X = 0).

b) Calculate Pr(X = 0) + Pr(X = 1) + Pr(X = 2) = 1 - Pr(X = 3) - Pr(X = 4).

3. Hello, Hazel!

From a lot of 10 missiles, 4 are selected at random and fired.
If the lot contains 3 defective missiles that will not fire,
what is the probability that:

a) all 4 will fire?

Let: . $\begin{array}{ccc} D &=& \text{de{f}ective} \\ G &=& \text{good} \end{array}$

There are: . ${10\choose4} \:=\:210$ possible samples.

There are: . ${7\choose4} \:=\:35$ samples with 4 G's.

Therefore: . $P(\text{4 G}) \:=\:\frac{35}{210} \:=\:\frac{1}{6}$

b) at most 2 will not fire ?

The opposite of "at most 2 D's" is "3 D's (and 1 G)".

The number of samples with 3 D's and 1 G is: . ${3\choose3}{7\choose1} \:=\:7$

Hence: . $P(\text{3 D}) \:=\:\frac{7}{210} \:=\:\frac{1}{30}$

Therefore: . $P(\text{at most 2 D}) \;=\;1 - \frac{1}{30} \:=\:\frac{29}{30}$

4. Originally Posted by Soroban
Hello, Hazel!

The opposite of "at most 2 D's" is "3 D's (and 1 G)".

The number of samples with 3 D's and 1 G is: . ${3\choose3}{7\choose1} \:=\:7$

Hence: . $P(\text{3 D}) \:=\:\frac{7}{210} \:=\:\frac{1}{30}$

Therefore: . $P(\text{at most 2 D}) \;=\;1 - \frac{1}{30} \:=\:\frac{29}{30}$

But for at most 2 will fire :
means 1G, 3D and also 2G 2D right ?