Two fair coins are tossed simultaneously. Find the probability of getting:
(a) 2 tails
(b) no head
(c) no tail
(d) same face on each coin
(e) different faces on the coin
Answers
. Two different coins are tossed randomly. Find the probability of:
(i) getting two heads
(ii) getting two tails
(iii) getting one tail
(iv) getting no head
(v) getting no tail
(vi) getting at least 1 head
(vii) getting at least 1 tail
(viii) getting atmost 1 tail
(ix) getting 1 head and 1 tail
Solution:
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.
probability of two tails = (1/2)×(1/2) = 1/4
probability of no head mean probability of two tails = 1/4
probability of no tail = probability of two head = (1/2)×(1/2) = 1/4
same face on each coin = either two heads or either two tails = 1/4 + 1/4 = 1/2
different face of coins = 1 - same face on each coin = 1 -1/2 = 1/2