Class 11th

14. The permutation of having even number at the last digit from the given 6 different digits namely 1, 2, 3, 4, 5, 6 to form a 3-digit number is

After taking one of the even number as last digit we can rearrange the remaining 5 digits taking 2 at a time. i.e.

Therefore, The required number = 20 * 3 = 60

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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

13. For every four-digit number we have to count the permutation of 10 digits namely 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9 taken 4 at a time

However, these permutation will include those where 0 is at 1000's place.

So, fixing 0 at 1000's place and rearranging the remaining 9 digits taking 3 at a time.

Therefore, The required number = ¹⁰P₄ – ⁹P₃

= 5040 – 504

= 4536 ways

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.