Question

A bag contains 20 chocolate balls of two flavours (milk and dark), and two wrapper colours (blue and red). There are 8 dark chocolate balls and 7 milk chocolate balls with blue wrappers, and 2 dark chocolate balls and 3 milk chocolate balls with red wrappers. A student randomly selects a chocolate ball from the bag. Let 'milk' be the event that the selected chocolate ball is a milk chocolate ball, and 'red wrapper' be the event that the selected chocolate ball has a red wrapper. Assume the chocolate balls are indistinguishable from one another during selection. Which of the following statements is true?

a. P( milk )

b. P( milk )=P( milk ∣ red wrapper ).
c. P( milk )>P( milk ∣ red wrapper ).

Answer 1

The probability of selecting a milk chocolate ball is 0.5, and the probability of selecting a milk chocolate ball given that it has a red wrapper is 0.6. Therefore, the true statement is c: P(milk) < P(milk | red wrapper).

Step-by-step explanation:

To solve the problem, we will calculate the probability (P) of drawing a milk chocolate ball from the bag and the probability of drawing a milk chocolate ball given that we have a red wrapper. The total number of chocolate balls is 20.

P(milk) is calculated by dividing the total number of milk chocolate balls by the total number of chocolate balls in the bag. There are 7 milk chocolate balls with blue wrappers and 3 with red wrappers, making 10 milk chocolate balls in total. So, P(milk) = 10 / 20 = 0.5.

P(milk | red wrapper) is the probability of drawing a milk chocolate when we know that a red wrapper has been selected. There are 2 dark and 3 milk chocolate balls with red wrappers, making 5 total red-wrapped balls. P(milk | red wrapper) = 3 / 5 = 0.6.

By looking at the numbers, we see that P(milk) is less than P(milk | red wrapper). Hence, statement c is true: P(milk) < P(milk | red wrapper).