Answer to Question #186893 in Statistics and Probability for MiyaMizu

Question

. From a box of a dozen rocket-propelled grenade (RPG), 4 are selected at random and fired. If the box contains 3 defective RPG’s, what is the probability that (a) all 4 will fire? (b) at most 2 will not fire?m

Answer 1

Solution:

Given, $n=4,p=\frac9{12}=\frac34,q=\frac14$

Here, p represents the success of RPG will fire.

$X\sim Bin(n,p)$

(a): $P(X=4)=^4C_4(\frac3{4})^4(\frac1{4})^0=0.3164$

(b): $P(X\le2)=1-P(X>2)=1-[P(X=3)+P(X=4]$

$=^4C_3(\frac3{4})^3(\frac1{4})^1+^4C_4(\frac3{4})^4(\frac1{4})^0$

$=0.7382$

Now, the probability that at most two will not fire$=1-0.7382=0.2618$