Probability

47. Since the die has two faces each mark 1, three faces each marks 2 and one face mark 3.

The possible sample space of outcome is.

S = {1, 2, 3} so, n (s) = 6

(i) P (2) $\frac{3}{6}=\frac{1}{2}$

(ii) P (1 or 3) = P (1) + P (3) = $\frac{2}{6}+\frac{1}{6}=\frac{2+1}{6}=\frac{3}{6}=\frac{1}{2}$

(iii) P (not 3) = 1 P (3) = 1 -$\frac{1}{6}$ = $\frac{6-1}{6}=\frac{5}{6}$

46. Total number of ways of drawing 4 cards from a duck of 52 cards,  n (s) = ⁵²C₄

Total no. of diamond cards = 13

Similarly, total. no. of spades cards = 13

45. Given,

No. of red marbles= 10

No. of blue marbles = 20

No. of green marbles = 30.

So, total no. of marbles = 10 + 20 + 30 = 60

Now, we are to select 5 marbles from the given 60 marbles.

So the sample space is:

n (s) = ⁶⁰c₅.

44. Given that, total number of student, n (S) = 60

Let A: student opted for NCC

n (A) = 30

B: student opted for NSS

n (B) = 32

And student who opted both NCC and NSS, n (A∩B) = 24

(i) Probability that student opted for NCC or NSS,

P (A∪B) = P (A) + P (B) – P (A∩B)

$=\frac{n\left(\right)}{n\left(\right)}+\frac{n\left(\right)}{n\left(\right)}-\frac{n(A\cap B)}{n\left(\right)}$

$=\frac{30}{60}+\frac{32}{60}-\frac{24}{60}$

$=\frac{30+32-24}{60}=\frac{38}{60}=\frac{19}{30}$

(ii) Probability that student opted neither NCC or NSS

P (not A and not B) = P (A'∩B') = P (A∪B)' = 1 – P (A∪B)

$=1-\frac{19}{30}=\frac{30-19}{30}=\frac{11}{30}$

(iii) Probabilities that student opted NSS but not NCC

P (B but not A) = P (B) – P (A∩B)

$=\frac{32}{30}-\frac{24}{30}$

$=\frac{32-24}{30}=\frac{8}{30}=\frac{2}{15}$

43. Let A: student passing in Hindi

B: student passing in English

Given, P (B) = 0.75

P (A∩B) = 0.5, passing both subject

And P (A'∩B') = 0.1, i.e., passing neither subject

P (A∪B)' = 0.1

1 – P (A∪B) = 0.1

P (A∪B) = 1 – 0.1

P (A∪B) = 0.9

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

P (A) = P (A∪B) + P (A∩B) – P (B)

P (A) = 0.9 + 0.5 – 0.75 = 0.65

? The probability of passing Hindi examination is 0.65.

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)

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

$=\frac{3}{8}$