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.

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.

it's equivalent, so we can say that its equal.

let's call the denominator x and the numerador is 18 less then denominator, x - 18.

multiply

2.x = 5.(x - 18)

2x = 5x - 90

90 = 5x - 2x

90 = 3x

x = 90/3

x = 30

so: