Class 11th

42. Let A: Student passes 1st examination

So, P (A) = 0.8

And B: Student passes 2nd examination

So, P (B) = 0.7

Also probability of passing at least one examination is P (A∪B) = 0.95

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

0.95 = 0.8 + 0.7 – P (A∩B)

P (A∩B) = 0.8 + 0.7 – 0.95

P (A∩B) = 0.55

Hence, probability of passing both examination is 0.55.

41. Given that, 40% study Mathematics, 30% study Biology and 10% study both Mathematics and Biology.

Let A: Students study Mathematics.

P (A) = 40% = $\frac{40}{100}$ .

Let B: Students study Biology.

P (B) = 30% = $\frac{30}{100}$ .

So, P (A∩B) i.e. probability of student studying both Mathematics and Biology is

P (A∩B) = 10% = $\frac{10}{100}$

? P (A∪B); probability of student studying Mathematics or Biology is

P (A∪B) = P (A) + P (B) – P (A∩B)

$=\frac{40}{100}+\frac{30}{100}-\frac{10}{100}$

$=\frac{60}{100}$

$=\frac{3}{5}$ .

40. Given P (A) = 0.42

P (B) = 0.48

P (A∩B) = 0.16

(i) P (not A) = P (A') = 1 – P (A) = 1 – 0.42 = 0.58

(ii) P (not B) = P (B') = 1 – P (B) = 1 – 0.48 = 0.52

(iii) P (A or B) = P (A∪B) = P (A) + P (B) – P (A∩B)

= 0.42 + 0.48 – 0.16

= 0.74

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) = 0.35

P (A∩B) =?

P (A∪B) = 0.7

So, P (A∪B) = P (A) + P (B) – P (A∩B)

0.7 = 0.5 + 0.35 – P (A∩B)

P (A∩B) = 0.5 + 0.35 – 0.7

P (A∩B) = 0.15

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, P (A∩B) < P (A) and P (AB) < P (B)

The given probabilities are consistently defined.

34. 

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

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

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

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) = $\frac{7}{13}$ .