Question

In a box containing 100 bulbs, 10 are defective. The probability that out of a sample of 5 bulbs, none is defective is
(A)$10^{-1}$
(B)$\biggr\lgroup\frac{1}{2}\biggr\rgroup ^5$
(C)$\biggr\lgroup\frac{9}{10}\biggr\rgroup ^5$
(D)$\frac{9}{10}$

Answer 1

Answer:

is c. (9/10)^5

Step-by-step explanation:

Given in a box containing n=100 bulbs, 10 are defective. We need to calculate the probability that out of a sample of (n=5)

bulbs, none is defective. Let the number of non-defective bulbs in this sample be X.

It is a Bernoulli trial as they satisfy the conditions (i) finite number of trials, (ii) independent trials, (iii) there is a definite outcome and (iv) the probability of success does not change for each trial..

P (getting a defective bulb) =p=10/100=1/10→

q=1−p=9/10

Since X has a bionomial distribution, the probability of x success in n-Bernoulli trials, P(X=x)=nCx.px.qn–x where x=0,1,2,...,n and (q=1–p)

P (none of bulbs are defective) =P(X=0)=5C0.1/100.9/105–0=(9/10)5

Answer 2

The probability that out of a sample of 5 bulbs , none are defective is (C) (9/10)⁵.

• Let the event(X) be the number of defective bulbs.
• Picking bulbs is a Bernoulli's trial

• So X has a binomial distribution as :

P(X=x) = $^nC_xq^{n-x}p^x$

where n is the number of times we pick a bulb

p is probability of picking a defective bulb and

p+q = 1

• Now substituting in the formula , we find the value of n, q and p

n = 5

p = 10/100 = 1/10

q = 1-q =  1 - 1/10 = 9/10

• Now the distribution becomes,

P(X=x) = $^5C_x(\frac{9}{10})^{5-x}(\frac{1}{10})^x$

• Now, we need to find the probability that no bulb is defective

Therefore, x=0

• Now,

P(X=0) = $^5C_0(\frac{9}{10})^{5}(\frac{1}{10})^0$

P(X=0) = 1 × $(\frac{9}{10})^{5}$ ×1

P(X=0) = $(\frac{9}{10})^{5}$

• The probability that out of 5 bulbs selected no bulb is defective is $(\frac{9}{10})^{5}$ .