Answer to Question #323814 in Statistics and Probability for Bless

Question

A shipment of 20 cameras includes 3 that are defective. What is the minimum number of cameras that must be selected if we require that P(at least 1 defective)≥ .8?

Answer 1

In this question, we need to find the minimum number of cameras (r) that must be selected if we require that

$P(At\ least\ 1\ defective)=1-P(0\ defective)\ge 0.8$

$P(0\ defective) \le 0.2$

Let Y is defined as a number of defective cameras.

Here, random variable follows a hypergeometric distribution, because we are selecting cameras from twenty cameras of which some are defectives and some are non-defectives.

A random variable Y is said to have a hypergeometric probability distribution if and only if

$p(y)=P(Y=y)=\frac{(r,y)x(N-r,n-y)}{(N,n)}$

$(r,y)=C^r_y$

Where y is an integer 0,1,2...n, subject to the restrictions

$n-y \le N-r$

N=6,n=3, r-?

Hence we can find the value of by substituting the value of r from 4 to 8 into equation (1), and we will check the value of probability whether it is equal to 0.2 or not. Then we can conclude the minimum number of cameras r.

If r=4

$P(Y=0)=\frac{(4,0)x(20-4,3-0)}{(20,3)}=\frac{C^4_0C^{16}_3}{C^{20}_4}=0.492$

If r=5

$P(Y=0)=\frac{C^5_0C^{15}_3}{C^{20}_3}=0.3991$

If r=6

$P(Y=0)=\frac{C^6_0C^{14}_3}{C^{20}_3}=0.3193$

If r=7

$P(Y=0)=\frac{C^7_0C^{13}_3}{C^{20}_3}=0.2508$

If r=8

$P(Y=0)=\frac{C^8_0C^{12}_3}{C^{20}_3}=0.193$

Hence is the minimum number that the probability $P(At\ least\ 1\ defective) \ge 0.8$

Answer:

The minimum number of cameras is 8.