Probability

1. Given,

$P(E)=0.6$

$P(F)=0.3$

$P(E\cap F)=0.2$

$P(E/F)=\frac{P(E\cap F)}{P(F)}=\frac{0.2}{0.3}=\frac{2}{3}$

$P(F/E)=\frac{P(F\cap E)}{P(E)}=\frac{0.2}{0.6}=\frac{2}{6}=\frac{1}{3}$

54. Since, the lock can be open by a combination of four digits from the given ten digits I e, from 0 to 9. The number of ways of selecting 4 digits, = 10C₄

This combination of 4 digits can again be arranged within themselves in 41 ways. So, total number of

52. Total no. of person selected to represent the company n (s) = 5.

Let A: person is male.

A = {Harish, Rohan, Salim}

n (A) = 3

And B: person has one 35 yes of age.

B = {Sheetal, Salim}

n (B) = 2

And A ∩ B = {Salim}

n (A ∩ B) = 1

Probability that person is either male or over 35 years.

= P (A ∪ B) = P (A) + P (B) P (A ∩ B)

$=\frac{n(A)}{n(S)}+\frac{n(B)}{n(S)}-\frac{n(A\cap \; B)}{n(S)}$

= $=\frac{3}{5}+\frac{2}{5}-\frac{1}{5}$

$=\frac{3+2-1}{5}=\frac{4}{5}$

51. Given, P (A) = 0.54

P (B) = 0.69.

P (A ∩ B) = 0.35.

(i) P (A ∪ B) = P (A) + P (B) - P (A ∩ B)

= 0.54 + 0.69 - 0.35

= 0. 88

(ii) P (A? ∩ B? ) = P (A ∩ B)? = 1 - P (A ∪ B) = 1 - 0.88 = 0.12

(iii) P (A ∩ B? ) = P (A) - P (A ∩ B)

= 0.54 - 0. 35 = 0.19

(iv) P (B ∩ A? ) = P (B) - P (A ∩ B) = 0.69 - 0.35 = 0.34

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49. Here, out of 100 students, first section has 40 students and the rest I e, 60 students enters in second section.

As me and my friend are among the 100 students.

The no. of ways of selecting 2 students from the 100 students

= 100C₂

(a) When both enters first section if 2 of us are among the 40 students that are to be selected. Similarly, if both enters second section among the 60 students for that section.

(if 2 of us)

Hence, no. of ways of selecting both in same section = 40C₂ + 60C₂

Probability that both of us are in same section

48. Total no. of ticket for lottery sold = 10, 000

 No. of ticket that are awarded prize = 10

So, no. of ticket that are not awarded prize = 10.000 - 10

= 9990

(a) Now, probability of met getting a prize if we buy one ticket

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