Ncert Solutions Maths class 11th

31. When three coins are tosses we have the sample space,

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

So, n (S) = 8

(i) Let A: 3 heads occurs.

A = {HHH}

So, n (A) = 1

? P (A) = $\frac{1}{8}$

(ii) Let B: 2 heads occurs

B = {HHT, HTH, THH}

So, n (B) = 3

? P (B) = $\frac{3}{8}$

(iii) Let C: at least 2 heads occurs i.e. 2 heads or more

C = {HHT, HTH, THH, HHH}

So, n (C) = 4

? P (C) = $\frac{4}{8}=\frac{1}{2}$

(iv) Let D: at most 2 heads occurs i.e. 2 heads or less

D = {TTT, HTT, THT, TTH, HHT, HTH, THH}

So, n (D) = 7

? P (D) = $\frac{7}{8}$

(v) Let E: no head occurs

E = {TTT}

So, n (E) = 1

? P (E) = $\frac{1}{8}$

(vi) Let F: 3 tails occurs

F = {TTT}

So, n (F) = 1

? P (F) = $\frac{1}{8}$

(vii) Let G: exactly two tails occurs

G = {TTH, THT, HTT}

So, n (G) = 3

? P (G) = $\frac{3}{8}$ .

(viii) Let H: no tail occurs

H = {HHH}

So, n (H) = 1

? P (H) = $\frac{1}{8}$ .

(ix) Let I: atmost two tails i.e. two tails or less

I = {HHH, THH, HTH, HHT, HTT, THT, TTH}

So, n (I) = 7

? P (I) = $\frac{7}{8}$

30. When a coin is tossed four times we have the sample space,

S = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, THHT, TTHH, THTH, HTHT, HTTH, TTTH, TTHT, THTT, HTTT, TTTT}

So, n (S) = 16.

Case I: When the outcome is all head, the amount is 1 + 1 + 1 + 1 =? 4 gain

Case II: When the outcome is 3 head and one tail, the amount is

1 + 1 + 1 – 1.50 = 3 – 1.50 =? 1.50 gain

Case III: When the outcome is 2 head and 2 tail, the amount is

1 + 1 – 1.50 – 1.50 = 2 – 3 =? 1 lose.

Case IV: When the outcome is 1 head and 3 tail, the amount is

1 – 1.50 – 1.50 – 1.50 = 1 – 4.50 =? 3.50 lose.

Case V: When the outcome is all tail, the amount is

–1.50 – 1.50 – 1.50 – 1.50 = –6 =? 6 lose.

Let A be event such that person wins ?4. Then,

A = {HHHH}

So, n (A) = 1

? P (A) = $\frac{1}{16}$

Let B: person wins ?1.5. Then,

B = {HHHT, HHTH, HTHH, THHH}

So, n (B) = 4

? P (B) = $\frac{4}{16}=\frac{1}{4}$

Let C: person loses ?1. Then,

C = {HHTT, THHT, TTHH, THTH, HTHT, HTTH}

So, n (C) = 6

? P (C) = $\frac{6}{16}=\frac{3}{8}$

Let D: person loses ?3.50. Then,

D = {HTTT, THTT, TTHT, TTTH}

So, n (D) = 4

? P (D) = $\frac{4}{16}=\frac{1}{4}$

Let E: person loses ?6. Then,

E = {TTTT}

So, n (E) = 1

? P (E) = $\frac{1}{16}$ .