Question

‘Eight Ball’ is a game played on a pool table with 15 balls numbered 1 to 15 and a ‘cue ball’ that is solid and white. Of the 15 numbered balls, eight are solid (non-white) coloured and numbered 1 to 8 and seven are striped balls numbered 9 to 15. The 15 numbered pool balls (no cue ball) are placed in a large bowl and mixed, then one ball is drawn out at random.

Based on the above information, answer the following questions.

1. What is the probability that the drawn ball bears the number 8?
2. What is the probability that the drawn ball bears an even number?
3. What is the probability that the drawn ball bears a number, which is a multiple of 3?
4. What is the probability that the drawn ball is solid coloured and bears an even number?

Answer 1

Given Data:

Total number of pool balls in the bowl = 15

Sample space: S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}

Total number of possible outcomes: n(S) = 15

i. Probability that the drawn ball bears the number 8:

Favourable outcome: {8}

Number of favourable outcomes: \[n(E_1) = 1\]

Formula: \[P(E_1) = \frac{n(E_1)}{n(S)}\]

Substitution and Calculation: \[P(\text{bearing number } 8) = \frac{1}{15}\]

ii. Probability that the drawn ball bears an even number:

Favourable outcomes: Even numbers from 1 to 15 are {2, 4, 6, 8, 10, 12, 14}

Number of favourable outcomes: \[n(E_2) = 7\]

Formula: \[P(E_2) = \frac{n(E_2)}{n(S)}\]

Substitution and Calculation: \[P(\text{bearing an even number}) = \frac{7}{15}\]

iii. Probability that the drawn ball bears a number which is a multiple of 3:

Favourable outcomes: Multiples of 3 from 1 to 15 are {3, 6, 9, 12, 15}

Number of favourable outcomes: \[n(E_3) = 5\]

Formula: \[P(E_3) = \frac{n(E_3)}{n(S)}\]

Substitution and Calculation: \[P(\text{bearing a multiple of }3) = \frac{5}{15} = \frac{1}{3}\]

iv. Probability that the drawn ball is solid coloured and bears an even number:

Solid coloured balls are numbered 1 to 8.

Even numbered balls among the solid ones: {2, 4, 6, 8}

Number of favourable outcomes: \[n(E_4) = 4\]

Formula: \[P(E_4) = \frac{n(E_4)}{n(S)}\]

Substitution and Calculation: \[P(\text{solid coloured and even}) = \frac{4}{15}\]