Maths

30. A fair coin is tossed four times, and a person win Re 1 for each head and lose Rs 1.50 for each tail that turns up.

From the sample space calculate how many different amounts of money you can have after four tosses and the probability of having each of these amounts.

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30. When a coin is tossed four times we have the sample space,

S = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, THHT, TTHH, THTH, HTHT, HTTH, TTTH, TTHT, THTT, HTTT, TTTT}

So, n (S) = 16.

Case I: When the outcome is all head, the amount is 1 + 1 + 1 + 1 =? 4 gain

Case II: When the outcome is 3 head and one tail, the amount is

1 + 1 + 1 – 1.50 = 3 – 1.50 =? 1.50 gain

Case III: When the outcome is 2 head and 2 tail, the amount is

1 + 1 – 1.50 – 1.50 = 2 – 3 =? 1 lose.

Case IV: When the outcome is 1 head and 3 tail, the amount is

1 – 1.50 – 1.50 – 1.50 = 1 – 4.50 =? 3.50 lose.

Case V: When the outcome is all tail, the amount is

–1.5

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30. When a coin is tossed four times we have the sample space,

S = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, THHT, TTHH, THTH, HTHT, HTTH, TTTH, TTHT, THTT, HTTT, TTTT}

So, n (S) = 16.

Case I: When the outcome is all head, the amount is 1 + 1 + 1 + 1 =? 4 gain

Case II: When the outcome is 3 head and one tail, the amount is

1 + 1 + 1 – 1.50 = 3 – 1.50 =? 1.50 gain

Case III: When the outcome is 2 head and 2 tail, the amount is

1 + 1 – 1.50 – 1.50 = 2 – 3 =? 1 lose.

Case IV: When the outcome is 1 head and 3 tail, the amount is

1 – 1.50 – 1.50 – 1.50 = 1 – 4.50 =? 3.50 lose.

Case V: When the outcome is all tail, the amount is

–1.50 – 1.50 – 1.50 – 1.50 = –6 =? 6 lose.

Let A be event such that person wins ?4. Then,

A = {HHHH}

So, n (A) = 1

? P (A) = $\frac{1}{16}$

Let B: person wins ?1.5. Then,

B = {HHHT, HHTH, HTHH, THHH}

So, n (B) = 4

? P (B) = $\frac{4}{16} = \frac{1}{4}$

Let C: person loses ?1. Then,

C = {HHTT, THHT, TTHH, THTH, HTHT, HTTH}

So, n (C) = 6

? P (C) = $\frac{6}{16} = \frac{3}{8}$

Let D: person loses ?3.50. Then,

D = {HTTT, THTT, TTHT, TTTH}

So, n (D) = 4

? P (D) = $\frac{4}{16} = \frac{1}{4}$

Let E: person loses ?6. Then,

E = {TTTT}

So, n (E) = 1

? P (E) = $\frac{1}{16}$ .

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Permutations and Combinations Exercise 6.3 Solutions

12. How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?

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12. The permutation of 9 different digits taken 4 at a time is given by

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11. Evaluate $\frac{n!}{(n - r)!},$ when (i) n = 6, r = 2 (ii) n = 9, r = 5.

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11. i. n = 6, r = 2

$\frac{6!}{(6 - 2)!}$

= $\frac{6!}{4!}$

= $\frac{6 * 5 * (4!)}{(4!)}$

= 30

ii. n = 9, r = 5

$\frac{9!}{(9 - 5)!}$

= $\frac{9!}{4!}$

= $\frac{9 * 8 * 7 * 6 * 5 * (4!)}{4!}$

= 9 * 8 * 7 * 6 * 5

= 15,120

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29. There are four men and six women on the city council. If one council member is selected for a committee at random, how likely is it that it is a woman?

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29. Number of women in the city council n (A) = 6

As there are four men and six women the total number of person in the sample space is 4 + 6 = 10.

So, n (S) = 10

P (A) = $\frac{n (A)}{n (S)} = \frac{6}{10} = \frac{3}{5}$

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10. If $\frac{1}{6!} + \frac{1}{7!} = \frac{x}{8!}$ , find x

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10. We have,

$\frac{1}{6!}$ + $\frac{1}{7!}$ = $\frac{x}{8!}$

=> $\frac{1}{6!}$ + $\frac{1}{7 * 6!}$ = $\frac{x}{8 * 7 * 6!}$

=> 1 + $\frac{1}{7}$ = $\frac{x}{8 * 7}$

=> $\frac{8}{7}$ = $\frac{x}{8 * 7}$

=>x = 8 * 8

=>x = 64

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28. A fair coin with 1 marked on one face and 6 on the other and a fair die are bothtossed. find the probability that the sum of numbers that turn up is (i) 3 (ii) 12

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28. The sample space of the experiment is

S = { (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5) (6, 6)}

So, n (S) = 12.

(i) Let E be event such that sum of numbers that turn up is 3. Then,

E = { (1, 2)}

So, n (E) = 1

P (E) = $\frac{n (E)}{n (S)} = \frac{1}{12}$ .

(ii) Let F be event such that sum of number than turn up is 12. Then,

F = { (6, 6)}

So, n (F) = 1

P (F) = $\frac{n (F)}{n (S)} = \frac{1}{12}$ .

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9. Compute $\frac{8!}{6! * 2!}$

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9. $\frac{8!}{6! * 2!}$ = $\frac{8 * 7 * (6!)}{(6!) * 1 * 2}$ = 4 * 7 = 28

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27. A card is selected from a pack of 52 cards.

(a) How many points are there in the sample space?

(b) Calculate the probability that the card is an ace of spades.

(c) Calculate the probability that the card is (i) an ace (ii) black card.

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27. (a) Since there are 52 cards in the sample space,

n (S) = 52.

So, there are 52 sample points.

(b) In a deck of 52 cards there are 4 ace cards of which only one is of spades.

Hence, if A be an event of getting an ace of spades.

n (A) = 1

So, P (A) = $\frac{n (A)}{n (S)} = \frac{1}{52}$ .

(c) (i) Let B be an event of drawing an ace. As there are 4 ace cards we have,

n (B) = 4

So, P (B) = $\frac{n (B)}{n (S)} = \frac{4}{52} = \frac{1}{13}$ .

(ii) Let D be an event of drawing black cards. Since there are 26 black cards we have,

n (D) = 26.

So, P (D) = $\frac{n (D)}{n (S)} = \frac{26}{52} = \frac{1}{2}$

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8. Is 3 ! + 4 ! = 7 ! ?

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8. L.H.S = 3! + 4!

= (1 * 2 * 3) + (1 * 2 * 3 * 4)

= 6 + 24

= 30

R.H.S = 7!

= 1 * 2 * 3 * 4 * 5 * 6 * 7

= 5040

As, L.H.S ≠ R.H.S

3! + 4! ≠ 7!

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Permutation and Combinations Ex 6.2 Solutions

7. Evaluate

(i) 8 !

(ii) 4 ! – 3!

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