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GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2508
Given Kudos: 1053
GPA: 3.39
If a coin has an equal probability of landing heads up or tails up eac
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18 Oct 2022, 06:56
18 Oct 2022, 06:56
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Question Stats:
66% (01:05) correct
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If a coin has an equal probability of landing heads up or tails up each time it is flipped , what is the probability that the coin will land heads up exectly twice in 3 consecutive flips ?
A. 0.125
B. 0.25
C. 0.375
D. 0.5
E. 0.666
A. 0.125
B. 0.25
C. 0.375
D. 0.5
E. 0.666
Re: If a coin has an equal probability of landing heads up or tails up eac
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22 Oct 2022, 07:49
22 Oct 2022, 07:49
1
Expert Reply
we know that we want 2 H and 1 T, so we have 3 possibilities
HHT
HTH
THH
find the probability of each and then add....
HHT= 1/2*1/2*1/2 = 1/8
HTH= 1/2*1/2*1/2 = 1/8
THH= 1/2*1/2*1/2 = 1/8
1/8+1/8+1/8= 3/8 or 0.375
HHT
HTH
THH
find the probability of each and then add....
HHT= 1/2*1/2*1/2 = 1/8
HTH= 1/2*1/2*1/2 = 1/8
THH= 1/2*1/2*1/2 = 1/8
1/8+1/8+1/8= 3/8 or 0.375
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If a coin has an equal probability of landing heads up or tails up eac
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22 Oct 2022, 23:41
22 Oct 2022, 23:41
We need to find On three consecutive flips of a coin, what is the probability that the coin will land heads up exactly twice in 3 consecutive flips
Coin is tossed 3 times => Total number of cases = \(2^3\) = 8
Lets solve the problem using two methods
Method 1:
Out of the 8 cases there are only three cases in which we get exactly two heads. HHT, HTH and THH.
=> Probability of getting exactly two heads = \(\frac{3}{8}\) = 0.375
So, Answer will be C
Method 2:
We have three places _ _ _ and we need to find the two places in which we can get a Head. We can do that in 3C2 ways
=> \(\frac{3!}{2! * 1!}\) = 3 ways
=> Probability of getting exactly two heads = Number of ways * P(H) * P(H) * P(T) = 3 * \(\frac{1}{2}\) * \(\frac{1}{2}\) * \(\frac{1}{2}\) = \(\frac{3}{8}\) = 0.375
So, Answer will be C
Hope it helps!
Watch the following video to learn How to Solve Probability with Coin Toss Problems
Coin is tossed 3 times => Total number of cases = \(2^3\) = 8
Lets solve the problem using two methods
Method 1:
Out of the 8 cases there are only three cases in which we get exactly two heads. HHT, HTH and THH.
=> Probability of getting exactly two heads = \(\frac{3}{8}\) = 0.375
So, Answer will be C
Method 2:
We have three places _ _ _ and we need to find the two places in which we can get a Head. We can do that in 3C2 ways
=> \(\frac{3!}{2! * 1!}\) = 3 ways
=> Probability of getting exactly two heads = Number of ways * P(H) * P(H) * P(T) = 3 * \(\frac{1}{2}\) * \(\frac{1}{2}\) * \(\frac{1}{2}\) = \(\frac{3}{8}\) = 0.375
So, Answer will be C
Hope it helps!
Watch the following video to learn How to Solve Probability with Coin Toss Problems
