Question

A golfer's bag contains 24 golf balls, 18 of which are ProFlight brand and the other 6 are DistMax brand. Find the probability that he reaches in his bag and randomly selects 5 golf balls and 4 of them are ProFlights and the other 1 is DistMax.

Answer 1

To find the probability, we use the concept of combinations.

Step-by-step explanation:

To find the probability that the golfer randomly selects 5 golf balls and 4 of them are ProFlights and the other 1 is DistMax, we need to use the concept of combinations. The total number of ways to select 5 golf balls from a bag containing 24 balls is C(24,5). The number of ways to select 4 ProFlight balls and 1 DistMax ball from their respective totals is C(18,4) * C(6,1). Therefore, the probability is given by:

P = (C(18,4) * C(6,1)) / C(24,5)

Answer 2

1. Assuming with replacement, the probability is 7.91%; or

2. Assuming without replacement, the probability is 8.64%

Step-by-step explanation:

Total number of golf balls = 24

Let Pr denotes probability, P denotes ProFlights, D denotes DistMax.

The probability of selecting 5 balls can be with or without replacement. Since the question is silent on this, the answers to the methods are provided as follows:

1. Assuming with replacement

Pr(4 P and 1 D) = (18/24) × (18/24) × (18/24) × (18/24) × (6/24)

= 0.75 × 0.75 × 0.75 × 0.75 × 0.25

= 0.0791 = 7.91%

2. Assuming without replacement

Here, we assume that 4 ProFlights are selected first before 1 DistMax is selected, and the probability is as follows:

Pr(4 P and 1 D) = (18/24) × (17/23) × (16/22) × (15/21) × (6/20)

= 0.7500 × 0.7391 × 0.7273 × 0.7143 × 0.3000

= 0.0864 = 8.64%

Therefore, the probability that he reaches in his bag and randomly selects 5 golf balls and 4 of them are ProFlights and the other 1 is DistMax is 7.91% assuming with replacement or 8.64% assuming without replacement.