Question

s = {HH, HT, TT, TH} , n(s) = 4 find the trial, outcomes, sample space, event.

Answer 1

Answer:

When two different coins are tossed randomly, the sample space is given by

S = {HH, HT, TH, TT}

Therefore, n(S) = 4.

(i) getting two heads:

Let E1 = event of getting 2 heads. Then,

E1 = {HH} and, therefore, n(E1) = 1.

Therefore, P(getting 2 heads) = P(E1) = n(E1)/n(S) = 1/4.

(ii) getting two tails:

Let E2 = event of getting 2 tails. Then,

E2 = {TT} and, therefore, n(E2) = 1.

Therefore, P(getting 2 tails) = P(E2) = n(E2)/n(S) = 1/4.

(iii) getting one tail:

Let E3 = event of getting 1 tail. Then,

E3 = {TH, HT} and, therefore, n(E3) = 2.

Therefore, P(getting 1 tail) = P(E3) = n(E3)/n(S) = 2/4 = 1/2

(iv) getting no head:

Let E4 = event of getting no head. Then,

E4 = {TT} and, therefore, n(E4) = 1.

Therefore, P(getting no head) = P(E4) = n(E4)/n(S) = ¼.

(v) getting no tail:

Let E5 = event of getting no tail. Then,

E5 = {HH} and, therefore, n(E5) = 1.

Therefore, P(getting no tail) = P(E5) = n(E5)/n(S) = ¼.

(vi) getting at least 1 head:

Let E6 = event of getting at least 1 head. Then,

E6 = {HT, TH, HH} and, therefore, n(E6) = 3.

Therefore, P(getting at least 1 head) = P(E6) = n(E6)/n(S) = ¾.

(vii) getting at least 1 tail:

Let E7 = event of getting at least 1 tail. Then,

E7 = {TH, HT, TT} and, therefore, n(E7) = 3.

Therefore, P(getting at least 1 tail) = P(E2) = n(E2)/n(S) = ¾.

(viii) getting atmost 1 tail:

Let E8 = event of getting atmost 1 tail. Then,

E8 = {TH, HT, HH} and, therefore, n(E8) = 3.

Therefore, P(getting atmost 1 tail) = P(E8) = n(E8)/n(S) = ¾.

(ix) getting 1 head and 1 tail:

Let E9 = event of getting 1 head and 1 tail. Then,

E9 = {HT, TH } and, therefore, n(E9) = 2.

Therefore, P(getting 1 head and 1 tail) = P(E9) = n(E9)/n(S)= 2/4 = 1/2.