Two friends, Bob and Ben, each buy one ticket for a drawing. They can choose six numbers from a total of one hundred numbers (0 - 99) for their ticket. Bob chooses the numbers 1, 2, 3, 4, 5, 6, and Ben chooses the numbers 39, 45, 66, 72, 74, 89. Who has a higher probability of winning the drawing?

Given that Bob and Ben are two friends who buy one lottery ticket each. Each ticket contains six numbers from a total of one hundred numbers from 0 to 99.

Bob chooses the numbers 1, 2, 3, 4, 5, 6, and Ben chooses the numbers 39, 45, 66, 72, 74, 89.

We are to select the friend who has a higher probability of winning.

Let, 'S' be the sample space for the experiment.

Then, S = {0, 1, 2, . . . , 98, 99} ⇒ n(S) = 100.

Let, B and F are the events that Bob win and Ben win respectively.

Then,

B = {1, 2, 3, 4, 5, 6} ⇒ n(B) = 6

and

F = {39, 45, 66, 72, 74, 89} ⇒ n(F) = 6.

Therefore, the probability of even t B is

[tex]P(B)=\dfrac{n(B)}{n(S)}=\dfrac{6}{100}=6\%[/tex]

and the probability of event F is

[tex]P(B)=\dfrac{n(F)}{n(S)}=\dfrac{6}{100}=6\%[/tex]

Since, both the events have equal probabilities, so there is an equal chance of winning of Bob and Ben.

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