Question

There are 6 white and 3 orange ping-pong balls in a brown paper bag. Two balls are randomly chosen. Enter your answers as fractions. (The "Preview" simply displays your answer in nice mathematical text. It does not mean that your answer is either right or wrong.) a) How many total balls are in the bag? Preview b) What is the probability that the 1st ball is orange? P(1st = orange) = Preview c) What is the probability that the 2nd ball is also orange, given that the 1st ball was orange? P(2nd = orange | 1st = orange) = Preview d) What is the probability that both the 1st and the 2nd balls are orange? Preview Get help: Video

Answer 1

Answer:

a. 9

b. $\mathbf{\frac{1}{3}}$

c. $\mathbf{\frac{1}{4}}$

d. $\mathbf{\frac{1}{12}}$

Step-by-step explanation:

a. The total number of balls in the bag is

$6 + 3 = 9$

b. Let A denote the event that the 1st chosen ball is orange. Let B denote the event that the 2nd chosen ball is orange.

$P(A) = \frac{\text{no. of orange balls in the bag}}{\text{Total no. of balls}} = \frac{3}{9} = \mathbf{\frac{1}{3}}$

c. If the first choice is orange, (i.e A occurs) then there are 9 - 1 = 8 balls left in the bag with 3 - 1 = 2 orange balls.

$P(B|A) = \frac{2}{8} = \mathbf{\frac{1}{4}}$

d. Here we are required to compute $P(B \cap A)$. We use the fact that

$P(B|A) = \frac{P(B \cap A)}{P(A)} \implies P(B \cap A) = P(B | A) \times P(A) = \frac{1}{4} \times \frac{1}{3} = \mathbf{\frac{1}{12}}$