Two coins are tossed simultaneously and we get two heads 150 times, one head 240 times and no head 2210 times. find the probability of getting i) no head ii) no tail
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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.
The solved examples involving probability of tossing two coins will help us to practice different questions provided in the sheets for flipping 2 coins.
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