Three coins are tossed once. Find the probability of getting
(i) 3 Heads (ii) 2 Heads (iii)Atleast 2 Heads
(iv) Atmost 2 Heads (v) No Head (vi) 3 Tails
(vii) Exactly 2 Tails (viii) No Tail (ix) Atmost two tails

Solution:

When three coins are tossed once, the sample space is given by

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

Hence, n (S) = 8

It is known that the probability of an event A is given by

P (A) = Number of outcomes favourable to A/Total number of possible outcomes

= n (A)/n (S)

(i) Let B be the event of the occurrence of 3 heads.

Accordingly, B = {HHH}

= n (B)/n (S)

= 1/8

(ii) Let C be the event of the occurrence of 2 heads.

Accordingly, C ={HHT , HTH ,THH}

P (C) = n (C)/n(S)

= 3/8

(iii) Let D be the event of the occurrence of at least 2 heads.

Accordingly, D = {HHH, HHT, HTH, THH}

P (D) = n (D)/n(S)

= 4/8

= 1/2

(iv) Let E be the event of the occurrence of atmost 2 heads.

Accordingly, E = {HHT, HTH, THH, HTT, THT, TTH, TTT}

P (E) = n (E)/n(S)

= 7/8

(v) Let F be the event of the occurrence of no

Accordingly, F = {TTT}

P (F) = n (F)/n(S)

= 1/8

(vi) Let G be the event of the occurrence of 3 tails.

Accordingly, G = {TTT}

P (G) = n (G)/n(S)

= 1/8

(vii) Let H be the event of the occurrence of exactly 2 tails.

Accordingly, H = {HTT, THT, TTH}

P (H) = n (H)/n(S)

= 3/8

(viii) Let I be the event of the occurrence of no tail.

Accordingly, I = {HHH}

P (I) = n (I)/n(S)

= 1/8

(ix) Let J be the event of the occurrence of at most 2 tails.

Accordingly, J = {HHH, HHT, HTH, THH, HTT, THT, TTH}

P (J) = n (J)/n(S)

= 7/8

NCERT Solutions Class 11 Maths Chapter 16 Exercise 16.3 Question 8