Question

Question B2 (a) The table below shows the discrete probability distribution of a random variable X: X P(X) 1 0. 2 7 18 W Tk 25 0.1 Suppose u = 12. Find the values of w and k. Hence, find the variance. (5 marks) (b) Suppose 15 students in a class are to be assigned randomly into 5 groups for group presentation. Each group contains 3 students. In how many different ways can these groups be formed? (3 marks) Suppose 5 cards are randomly drawn from an ordinary deck of playing 52 cards one-by- one without replacement. Find the probability of getting at most one Ace card. (Correct your final answer to 4 decimal places) (4 marks)

Answer 1

(a) 0.2 + w + k + 0.1 = 1 (since all probabilities must add up to 1)

Hence, w + k = 1-0.2 - 0.1 = 0.7

w = 0.7 - k

Mean = 1 x 0.2 + 7 x w + 18 x k + 25 x 0.1 = 12

=> 7w + 18k = 12 - 0.2 - 2.5 = 9.3

Replacing the value of w as 0.7-k in the above equation, we have

7 x (0.7-k) + 18k = 9.3

4.9 - 7k + 18k = 9.3

11k = 9.3 - 4.9 = 4.4

k = 4.4/11 = 0.4

w = 0.7-k = 0.7-0.4 = 0.3

Variance = 0.2 x (1-12)² + 0.3 x (7-12)² + 0.4 x (18-12)² + 0.1 x (25-12)² = 63

(b). for the first group, 3 students can be selected from 15 in C[15,3] ways.

Then for the 2nd group, 3 students can be selected from the rest 12 students in C[12,3] ways.

For the 3rd group, 3 students can be selected from the balance 9 students in C[9,3] ways and so on.

So all possible ways = C[15,3] x C[12,3] x C[9,3] x C[6,3] x C[3,3]However, the above possible combinations also contains scenarios where identical students in a single group for either group 1 or 2 or..so on till group. In other words, order of the groups is not important. Such combinations are 5!

Hence, number of distinct possible groups =

$\frac{\binom{15}{3}*\binom{12}{3}*\binom{9}{3}*\binom{6}{3}*\binom{3}{3}}{5!} = \frac{15!}{5!*3!*3!*3!*3!*3!}=1401400$

(c)There are four Aces, so 48 non-Aces

Probability of getting 0 aces in a random draw = $\frac{\binom{48}{5}}{\binom{52}{5}}$

Probability of getting one Ace in a random draw = $\frac{\binom{48}{4}*\binom{4}{1}}{\binom{52}{5}}$

Probability of getting at most one Ace = $\frac{\binom{48}{5}+\binom{48}{4}*\binom{4}{1}}{\binom{52}{5}} =0.9583$