Answer:
1
Explanation:
Given:-
- The box has n = 20 light-bulbs
- The number of defective bulbs, d = 5
Find:-
what's the probability that at most two of them are defective
Solution:-
- We will pick 2 bulbs randomly from the box. We need to find the probability that at-most 2 bulbs are defective.
- We will define random variable X : The number of defective bulbs picked.
Such that, P ( X ≤ 2 ) is required!
- We are to make a choice " selection " of no defective light bulb is picked from the 2 bulbs pulled out of the box.
- The number of ways we choose 2 bulbs such that none of them is defective, out of 20 available choose the one that are not defective i.e n = 20 - 5 = 15 and from these pick r = 2:
X = 0 , Number of choices = 15 C r = 15C2 = 105 ways
- The probability of selecting 2 non-defective bulbs:
P ( X = 0 ) = number of choices with no defective / Total choices
= 105 / 20C2 = 105 / 190
= 0.5526
- The number of ways we choose 2 bulbs such that one of them is defective, out of 20 available choose the one that are not defective i.e n = 20 - 5 = 15 and from these pick r = 1 and out of defective n = 5 choose r = 1 defective bulb:
X = 1 , Number of choices = 15 C 1 * 5 C 1 = 15*5 = 75 ways
- The probability of selecting 1 defective bulbs:
P ( X = 1 ) = number of choices with 1 defective / Total choices
= 75 / 20C2 = 75 / 190
= 0.3947
- The number of ways we choose 2 bulbs such that both of them are defective, out of 5 available defective bulbs choose r = 2 defective.
X = 2 , Number of choices = 5 C 2 = 10 ways
- The probability of selecting 2 defective bulbs:
P ( X = 2 ) = number of choices with 2 defective / Total choices
= 10 / 20C2 = 10 / 190
= 0.05263
- Hence,
P ( X ≤ 2 ) = P ( X =0 ) + P ( X = 1 ) + P (X =2)
= 0.5526 + 0.3947 + 0.05263
= 1