Question

One day, Karen bought 23 random cans of soup from a grocery store. Suppose that 6% of cans sold at that particular grocery store are dented, and assume the store's inventory is large enough that no individual customer's purchase affects the dent rate for the remaining cans. What is the probability that Karen has bought at least one dented can? Express your answer as a percentage precise to one decimal place.
Please explain.

Answer 1

Answer: 75.9%

Step-by-step explanation:

Given : Number of cans bought by Karen : n=23

Proportion of cans sold at that particular grocery store are dented : p= 0.06

We assume that the store's inventory is large enough that no individual customer's purchase affects the dent rate for the remaining cans.

Let x be binomial variable that represents the dented can.

Using binomial probability formula ,

$P(x)=^nC_xp^x(1-p)^{n-x}$

The probability that Karen has bought at least one dented can:-

$p(x\geq1)=1-P(x=0)\\\\=1-^{23}C_{0}(0.06)^0(0.94)^{23}\\\\=(1)(0.94)^{23}\\\\=1-0.240957602184=0.759042397816\approx0.7590=75.9\%$

Hence, the required probability = 75.9%

Answer 2

Answer:

P(X=1) = 0.338

Step-by-step explanation:

Given data:

Total cans, n = 24,

percentage of cans sold p = 0.06

Let X denote number of cans that are dented.

So,  we have following relation from binomial distribution

$P(X = x) = ^nC_x (p^x)(1-p)^{n-x}$

Putting values we get:

$P(X = x) = ^{23}C_x(0.06^x)(1- 0.03)^{23-x}$

Probability of having one purchase car dented i.e x = 1

$P(X=1) = ^{23}C_1 (0.06^1)(1-0.06)^{23 -1}$

After solving we get:

P(X=1) = 0.338