Question

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

i) All 4 will fire.

ii) At most 2 will not fire.

iii) How many defective missiles might we expect to be included among the 4 that are selected.

Answer 1

i) The number of such chooses is the number of chooses of $4$ missles from the $7$ good ones, which equals $\binom{7}{4}$. The total number of different chooses is $\binom{10}{4}$, so the probability that all chosen missles will shoot equals $\binom{7}{4}/\binom{10}{4}=1/6$ .

ii) $P(\text{at most }2 \text{ will not fire})=1-P(3\text{ missles will not fire})$. The number of chooses of $4$ missles such that there are $3$ defective ones between them is $7$ ($3$ defected + an arbitrary missle from the left), so $P(\text{at most }2 \text{ will not fire})=1-7/\binom{10}{4}=29/30$.

iii) Number the defective missles, and let $X_i=1$ if the $i$-th missle is chosen and $X_i=0$ if not. Then, by linearity of expectation, the expected number of defective missles chosen equals

$E(X_1+X_2+X_3)=E(X_1)+E(X_2)+E(X_3)=P(X_1\text{ is chosen})+P(X_2\text{ is chosen})+P(X_3\text{ is chosen}).$

The number of choises such that the $i$-th defective missle is chosen is $\binom{9}{4}$ (we choose this missle and $3$ arbitrary from the $9$ missles left), so

$E(X_1+X_2+X_3)=3\cdot\binom{9}{4}/\binom{10}{4}=6/5.$