QuIZ I TIME 15 Hours Contribution to Final 15% 1 A random sample of 25 people re

Question

QuIZ I TIME 15 Hours Contribution to Final 15% 1 A random sample of 25 people recorded the number of glasses of water they drank in panicular week. The results are shown below 19 32 14 29. 26 36 45 4 47 28 17 38 1 46 18 26 22 19 2 0 0) Draw a stom-and-leaf diagram to represent the data4 ) On graph paper drasw a box-and-whisker plo to represent the dats 1o) 2A commites of 6 people is to be chosen from 5 men and women In how many wys can this be done 0) If there are more women than men on the commine 4 141 If the comminee consists of 3 men and 3 women but two particular men refluse to be on the commitnee together?14 ne panticular commitoe consists of S women and 1 man ) n how many different ways can the comminee be arranged in a line if the man is not at either end (4 eto incr company A, 30% to B and 20% to C. Atau from company A arrives late 4 % the time, a taxi from company B arrives late 6% of the time and tau from company hen Alex needs a taxi, he rings one of three taxi companies, A, B or C. 50% ofhis calls late 17% of the time. () Find the probability that, when Alex rings for a taxi, it arives late. (9 ) Given that Alex's taxi arives late, find the conditional probability thet rives rang company B 4

Explanation / Answer

Question on committee formation

We have 6 men and 6 women from amongst whom we need to form a committee of 6.

This can be done in 12C6 = (12!)/{(6!)(6!)} ways

= (12 x 11 x 10 x 9 x 8 x 7)/(6 x 5 x 4 x 3 x 2 x 1) = 924 ANSWER 1

Part (1)

Committee to have more men than women. This is possible in the following combinations:

Men

6

5

4

Women

0

1

2

#of committees

(6C6)( 6C0) = 1

(6C5)( 6C1) = 36

(6C4)( 6C2) = 25

Total number of committees = 62 ANSWER

Part (2)

Committee to have 3 men and 3 women, but two particular men refuse to work together

This is possible in the following combinations:

Case 1: Committee does not have either of those two particular men

So, 3 men must be out of the remaining 4 men and 3 women must be out of 6 women.

Number of possibilities = (4C3)( 6C3) = 4 x 20 = 80.

Case 2: Committee has just one of those two particular men

One of those two particular men can be selected in 2 ways. And then, 2 remaining men must be out of the remaining 4 men and 3 women must be out of 6 women.

Number of possibilities = 2 x (4C2)( 6C3) = 2 x 6 x 20 = 240.

Total number of committees = 320 ANSWER

Part (3)

Committee to have 1 man and 5 women.

This is possible in the following combinations:

(6C1)( 6C5) = 6 x 6 = 36 ANSWER

Part (4)

Members of the committee formed under Part (3) is to be arranged in a line so that the man is not at either end.

So, the man can occupy any one of the remaining 4 positions (leaving the two positions at the ends) in 4 ways. The remaining 5 women can occupy the remaining 5 positions in 5! Ways.

Thus, the total number of arrangements = 4 x 120 = 480 ANSWER

Men

6

5

4

Women

0

1

2

#of committees

(6C6)( 6C0) = 1

(6C5)( 6C1) = 36

(6C4)( 6C2) = 25

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