Ncert Solutions Maths class 11th

10. We have,

$\frac{1}{6!}$ + $\frac{1}{7!}$ = $\frac{x}{8!}$

=> $\frac{1}{6!}$ + $\frac{1}{7*6!}$ = $\frac{x}{8 *7*6!}$

=> 1 + $\frac{1}{7}$ = $\frac{x}{8*7}$

=> $\frac{8}{7}$ = $\frac{x}{8 *7}$

=>x = 8 * 8

=>x = 64

28. The sample space of the experiment is

S = { (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5) (6, 6)}

So, n (S) = 12.

(i) Let E be event such that sum of numbers that turn up is 3. Then,

E = { (1, 2)}

So, n (E) = 1

P (E) = $\frac{n(E)}{n(S)}=\frac{1}{12}$ .

(ii) Let F be event such that sum of number than turn up is 12. Then,

F = { (6, 6)}

So, n (F) = 1

P (F) = $\frac{n(F)}{n(S)}=\frac{1}{12}$ .

9.  $\frac{8!}{6! * 2!}$ = $\frac{8 *7*\left(6!\right) }{\left(6!\right)*1*2}$ = 4 * 7 = 28

27. (a) Since there are 52 cards in the sample space,

n (S) = 52.

So, there are 52 sample points.

(b) In a deck of 52 cards there are 4 ace cards of which only one is of spades.

Hence, if A be an event of getting an ace of spades.

n (A) = 1

So, P (A) = $\frac{n(A)}{n(S)}=\frac{1}{52}$ .

(c) (i) Let B be an event of drawing an ace. As there are 4 ace cards we have,

n (B) = 4

So, P (B) = $\frac{n(B)}{n(S)}=\frac{4}{52}=\frac{1}{13}$ .

(ii) Let D be an event of drawing black cards. Since there are 26 black cards we have,

n (D) = 26.

So, P (D) = $\frac{n(D)}{n(S)}=\frac{26}{52}=\frac{1}{2}$

8. L.H.S = 3! + 4!

= (1 * 2 * 3) + (1 * 2 * 3 * 4)

= 6 + 24

= 30

R.H.S = 7!

= 1 * 2 * 3 * 4 * 5 * 6 * 7

= 5040

As, L.H.S ≠ R.H.S

3! + 4! ≠ 7!

1. We know that, n! = n (n – 1) (n – 2)…….

i. 8!

8! = 1 * 2 * 3 * 4 * 5 * 6 * 7 * 8

= 40320

ii. 4! – 3!

= (1 * 2 * 3 * 4) – (1 * 2 * 3)

= 24 – 6

= 18

6. To generate a signal which requires 2 flags one below another we can have the following combination of any of the 5 flag at top and the one of the remaining 4 flag at the bottom.

Hence, total no. of possible combination = 5 * 4 = 20

25. When a coin is tossed twice we have the sample space

S = {TT, TH, HT, HH}

So, n (S) = 4

Let A be the event of getting at least one tail.

Then, A = {TH, HT, TT}

So, n (A) = 3

Therefore, P (A)= Numver of outcome favorable to A/Total possible outcomes = $\frac{}{}=\frac{n(A)}{n(S)}$ .

$=\frac{3}{4}$

5. When a coin is tossed one time a head or a tail is the possible outcome.

So, when a coin is tossed 3 times the total number of possible outcomes = 2 * 2 * 2 = 8