Probability

32. Let A be the event

Given that, P (A) = $\frac{2}{11}$

So, P (not A) = P (S) – P (A) = $1-\frac{2}{11}=\frac{11-2}{11}=\frac{9}{11}$

31. When three coins are tosses we have the sample space,

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

So, n (S) = 8

(i) Let A: 3 heads occurs.

A = {HHH}

So, n (A) = 1

? P (A) = $\frac{1}{8}$

(ii) Let B: 2 heads occurs

B = {HHT, HTH, THH}

So, n (B) = 3

? P (B) = $\frac{3}{8}$

(iii) Let C: at least 2 heads occurs i.e. 2 he

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 ta

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

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

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 dra

26. The sample space of throwing s dice is

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

(i) Let A be event such that a prime number will appear. Then,

A = {2, 3, 5}

? n (A) = 3

Here; P (A) = $\frac{n(A)}{n(S)}=\frac{3}{6}=\frac{1}{2}$

(ii) Let B be event such that a number greater than or equal to 3 will appear. Then

B = {3, 4, 5, 6}

So

25. When a coin is tossed twice we have the sample space

S = {TT, TH, HT, HH}

So, n (S) = 4

Let A be the event of getting at least one tail.

Then, A = {TH, HT, TT}

So, n (A) = 3

Therefore, P (A)= Numver of outcome favorable to A/Total possible outcomes = $\frac{}{}=\frac{n(A)}{n(S)}$ .

$=\frac{3}{4}$

24. (a) P (S) = P (W1) + P (W2) +P (W3) + P (W4) +P (W5) +P (W6) +P (W7)

P (S) = 0.1 + 0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6

P (S) = 1

As the probability of sample space is ‘one’ the given assignment of probabilities is valid.

(b) P (S) = P (W1) + P (W2) +P (W3) + P (W4) +P (W5) +P (W6) +P (W7)

P (S) =

 23. (i) A ∩B = { (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}∩ { (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),