Question

7. A box contains 12 white marbles and 5 black marbles. Suppose we randomly draw a marble from the box, replace it, and then randomly draw another marble from the box. (This means that we might observe the same marble twice). What is the probability that both the marbles are white? Write your answer as a decimal accurate to three decimal places.

8. Suppose that 7.3 % of the items produced by a factory are defective. If 5 items are chosen at random, what is the probability that none of the items are defective? Write your answer as a decimal accurate to three decimal places.

9. Suppose that 3.6 % of the items produced by a second factory are defective. If 5 items are chosen at random from the second factory, what is the probability that exactly one of the items is defective? Write your answer as a decimal accurate to three decimal places.

10. Suppose that 2.9 % of the items produced by a third factory are defective. If 5 items are chosen at random from the third factory, what is the probability that exactly two of the items are defective? Write your answer as a decimal accurate to three decimal places.

Answer 1

Question 7

We are given

12 white + 5 black = 17 marbles

We draw marbles with replacement.

So, Probability of drawing white marble = p = P(White) = 12/17 = 0.705882

Now, we have to find P(both marbles are white)

We have n = 2, p = 0.705882, q = 1 – p = 1 - 0.705882 = 0.294118

We have to find P(X=2)

P(X=x) = nCx*p^x*q^(n – x)

P(X=2) = 2C2*0.705882^2*0.294118^(2 – 2)

P(X=2) = 1*0.705882^2*0.294118^0

P(X=2) = 1*0.705882^2*1

P(X=2) = 0.705882^2

P(X=2) = 0.498269

Required probability = 0.498

Question 8

We are given n = 5, p = 0.073, q = 1 – p = 1 – 0.073 = 0.927

We have to find P(X=0)

P(X=x) = nCx*p^x*q^(n – x)

P(X=0) = 5C0*0.073^0*0.927^(5 – 0)

P(X=0) = 1*1*0.927^5

P(X=0) = 0.68454

Required probability = 0.685

Question 9

We are given n = 5, p = 0.036, q = 1 – p = 1 – 0.036 = 0.964

We have to find P(X=1)

P(X=x) = nCx*p^x*q^(n – x)

P(X=1) =5C1*0.036^1*0.964^(5 – 1)

P(X=1) = 5*0.036^1*0.964^4

P(X=1) = 0.155446

Required probability = 0.155

Question 10

We are given n = 5, p = 0.029, q = 1 – p = 1 – 0.029 = 0.971

We have to find P(X=2)

P(X=x) = nCx*p^x*q^(n – x)

P(X=2) =5C2*0.029^2*0.971^(5 – 2)

P(X=2) = 10*0.029^2*0.971^3

P(X=2) = 0.007699

Required probability = 0.008