Question

A girl and a boy each randomly grab a piece of candy from a bowl containing 9 pieces of chocolate, 7 fruit chews, 9 lollipops, and 9 peppermints. Find the probability that both events A and B will occur. Give your answer a reduced fraction. Event A: A girl grabs a lollipop Event B: A boy grabs a fruit chew

Answer 1

The probability that both events A and B will occur is 0.56

Step-by-step explanation:

It is given that,

• There are 9 pieces of chocolate, 7 fruit chews, 9 lollipops, and 9 peppermints in a bowl.
• The total number of pieces in the bowl = 9+7+9+9 = 34 pieces.
• A girl and a boy each randomly grab a piece of candy from a bowl.

Therefore, two events A and B will occur.

Event A ⇒ A girl grabs a lollipop :

P (to grab lollipop) = No.of lollipops / Total pieces

We know that, no.of lollipops = 9 and the total pieces = 34

P (to grab lollipop) = 9 / 34.

Event B ⇒ A boy grabs a fruit chew :

P (to grab fruit chew) = No.of fruit chews / Total pieces in bowl after event A.

We know that, no.of fruit chews = 7 and the total pieces in the bowl after grabbing one lollipop is 33 pieces.

P (to grab fruit chew) = 7 / 33

Probability that both events A and B will occur :

P (event A and B) = P (to grab lollipop) × P (to grab fruit chew)

⇒ 9/34 × 7/33

⇒ 63/1122

⇒ 21/374

⇒ 0.56

The probability that both events A and B will occur is 0.56

Answer 2

The probability that both the events will occur is $\frac{63}{1156}$

Explanation:

Chocolate = 9

Fruit chews = 7

Lollipops = 9

Peppermints = 9

Total candies = 9 + 7 + 9 + 9

                     = 34

Probability of grabbing a lollipop = $\frac{9}{34}$

Probability of grabbing a fruit chew = $\frac{7}{34}$

Probability that both the events will occur = $\frac{9}{34} X \frac{7}{34}$

                                                                  = $\frac{63}{1156}$

Therefore, the probability that both the events will occur is $\frac{63}{1156}$