Question

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

Answer 1

There is a lot contain 10 missiles and if the lot contains 4 defective missiles that will not fire then the number of missiles will fire is (10-4)=6.

Now, 3 missiles are selected at random.

X: random variable denoting that number of missiles will fire from the selected missiles.

[ Bionomial distribution:- The binomial distribution is frequently used to model the number of successes in a sample of size n drawn with replacement from a population of size N. For this problems N=10, n=3 and if we consider missiles will fire as a success than random variable X follow Bionomial distribution. ]

So, Probability of success i.e. probability of a randomly selected missiles from the lot of 10 missiles will fire is 6/10

[Note:- probability of a event = (number of outcomes favorable to the event/ size of sample space) ; Here size of sample space= total number of missiles =10 and number of outcomes favorable to the event = number of missiles will fire =6 . So the required probability is 6/10]

Hence, X~Bin(n=3, P=6/10).

if X~ Bin(n,p) then pdf of X is given by,

; OP<!, 220, 1-.n. So for this publem, . P(x=x) = (2) (taje linija at we want to find the probability that all will fise; P(x-2) = (*) (4.) '(- 9 0.216

by we want to find the proobability that at most 2 will not fire, This is the complementary case of there is no missile will fine so our poequined poubability is, 1- P(x = 0) [ poub of any evento 1 - popus of complementory of that event? 71- P(x=0) 31-(3) (10.13(-) 3. 2.1-0.064 = 0.936.

Alternatie: X: 8.v tenoting that number of missies will not fize and cie consider missires will not fire as a success. Then, X~Bin (3, 4) and we want to find the probabilit thank at most 2 will not fize: des P(x32) - 1 - P(x= 3) 5) 1-(3)(4) } 1-0.064 24 0.936