Consider the experiment of tossing a fair coin until two heads or two tails appear in succession.
(a) Describe the sample space.
(b) What is the probability that the experiment ends before the sixth toss?
(c) What is the probability that the experiment ends after an even number of tosses?
(d) Given that the experiment ends with two heads, what is the probability that the experiment ends before the sixth toss?
(e) Given that the experiment does not end before the third toss, what is the probability that the experiment does not end after the sixth toss?
Answers
Answer:
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Probability of Tossing Two Coins
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Here we will learn how to find the probability of tossing two coins.
Let us take the experiment of tossing two coins simultaneously:
When we toss two coins simultaneously then the possible of outcomes are: (two heads) or (one head and one tail) or (two tails) i.e., in short (H, H) or (H, T) or (T, T) respectively; where H is denoted for head and T is denoted for tail.
Therefore, total numbers of outcome are 22 = 4
The above explanation will help us to solve the problems on finding the probability of tossing two coins.
Worked-out problems on probability involving tossing or flipping two coins:
1. 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.
The solved examples involving probability of tossing two coins will help us to practice different questions provided in the sheets for flipping 2 coins.