Probability

39. Given, P (not E or not F) = 0.25

P (E'∪F') = 0.25

P (E∩F)' = 0.25

1 – P (E∩F) = 0.25

P (E∩F) = 1 – 0.25

P (E∩F) = 0.75≠ 0

Hence, E and F are not mutually exclusive events.

38. Given, P (E) = $\frac{1}{4}$

P (F) = $\frac{1}{2}$

P (E and F) = P (E∩F) = $\frac{1}{8}$

(i) P (E or F) = P (E∪F) = P (E) + P (F) – P (E∩F)

$=\frac{1}{4}+\frac{1}{2}-\frac{1}{8}$

$=\frac{2+4-1}{8}$

$=\frac{5}{8}$

(ii) P (not E and not F) = P (E'∩F') = P (E∪F)' = 1 – P (E∪F)

$=1-\frac{5}{8}=\frac{8-5}{8}$

$=\frac{3}{8}$

37. 

Given P (A) = $\frac{3}{5}$

P (B) = $\frac{1}{5}$

As A and B are mutually exclusive events,

P (A∩B) = 0

Hence, P (A∪B) = P (A) + P (B) – P (A∩B)

$=\frac{3}{5}+\frac{1}{5}-0$

$=\frac{4}{5}$

36. (i) Given P (A) = $\frac{1}{3}$

P (B) = $\frac{1}{5}$

P (A∩B) = $\frac{1}{15}$

So, P (A∪B) = P (A) + P (B) – P (A∩B)

$=\frac{1}{3}+\frac{1}{5}-\frac{1}{15}$

$=\frac{5+3-1}{15}$

P (A∪B) = $\frac{7}{15}$

(ii) Given P (A) = 0.35

P (B) =?

P (A∩B) = 0.25

P (A∪B) = 0.6

So, P (A∪B) = P (A) + P (B) – P (A∩B)

0.6 = 0.35 + P (B) – 0.25

P (B) = 0.6 – 0.35 + 0.25

P (B) = 0.5

(iii) Given P (A) = 0.5

P (B) = 0.3

35. Given P (A) = 0.5

P (B) = 0.7

And P (A∩B) = 0.6

As P (A∩B) > P (A) which is not possible.

The given probabilities are not consistently defined.

(ii) Given, P (A) = 0.5

P (B) = 0.4

And P (A∪B) = 0.8

So, P (A∪B) = P (A) + P (B) – P (A∩B)

0.8 = 0.5 + 0.4 – P (A∩B)

P (A∩B) = 0.5 + 0.4 – 0.8

P (A∩B) = 0.1

Hence

34. 

. Since 6 numbers are to be choosen as fixed from a set a given 20 number, the sample space is

$n\left(\right)=20C_6=\frac{20!}{6!\left(20-6\right)!}=\frac{20\times 19\times 18\times 17\times 16\times 15\times 14!}{2\times 3\times 4\times 5\times 6\times 14!}$

$=\frac{38760}{1}\times \frac{14!}{14!}=38760$

Let A: person wins the prize.

In order to win the prize the 6 number has to be correct i.e. all 6 of the number are to be choosen from fixed 6 numbers we have,

$n(S){=}^6{}_{\mathrm{C6}}=1$

? P (A) = $\frac{1}{38760}$ .

33.  The sample space of word is

S = {A, S, A, S, I, N, A, T, I, O, N}

So, n (S) = 13.

(i) Let A: word is a vowel

A = {A, I, A, I, O}

So, n (A) = 6

? P (A) = $\frac{6}{13}$

(ii) Let B: Word is a consonant

B = {S, N, T, N}

So, n (B) = 7

? P (B) = $\frac{7}{13}$ .

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