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Probability-A fair coin is flipped 5 times
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Updated on: 05 Oct 2022, 06:26
Updated on: 05 Oct 2022, 06:26
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62% (00:45) wrong based on 27 sessions
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A fair coin is flipped 5 times.
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
The probability of getting more heads than tails |
1/2 |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Originally posted by goneewindy on 10 Aug 2015, 12:07.
Last edited by Carcass on 05 Oct 2022, 06:26, edited 2 times in total.
Last edited by Carcass on 05 Oct 2022, 06:26, edited 2 times in total.
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Re: Probability-A fair coin is flipped 5 times
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05 Oct 2022, 06:23
05 Oct 2022, 06:23
1
Given that A fair coin is flipped 5 times.
Quantity A: The probability of getting more heads than tails
Method 1
As there are odd number of coin tosses so there are only two outcomes possible. Either the number of heads will be more or the number of tails will be more. So there is equal change of occurrence of both the outcomes.
=> P(Getting more heads than tails) = \(\frac{1}{2}\)
=> Quantity A = Quantity B = \(\frac{1}{2}\)
So, Answer will be C
Method 2
Coin is tossed 5 times => Total number of cases = \(2^5\) = 32
P(Getting more heads than tails) = (Cases in which we will get 3H + 4H + 5H) / 32
Cases in which we get 3H
Cases in which we get 3H can be found by putting 3 heads in any of the 5 slots _ _ _ _ _ _ _
This can be done in 5C3 ways = \(\frac{5!}{3!*(5-3)!}\) = \(\frac{5*4*3!}{3!*2!}\) = 10 ways
Cases in which we get 4H
Cases in which we get 4H is same as getting 1 Tail and this can be found by putting 1 tail in any of the 5 slots _ _ _ _ _ _ _
This can be done in 5 ways by putting T in any of the 5 slots => 5 ways
Cases in which we get 5H
This can be done in 1 way (HHHHH)
=> P(Getting more heads than tails) = \(\frac{10 + 5 + 1 }{ 32 }\) = \(\frac{16}{32}\) = \(\frac{1}{2}\)
=> Quantity A = Quantity B = \(\frac{1}{2}\)
So, Answer will be C
Hope it helps!
Watch the following video to learn How to Solve Probability with Coin Toss Problems
Quantity A: The probability of getting more heads than tails
Method 1
As there are odd number of coin tosses so there are only two outcomes possible. Either the number of heads will be more or the number of tails will be more. So there is equal change of occurrence of both the outcomes.
=> P(Getting more heads than tails) = \(\frac{1}{2}\)
=> Quantity A = Quantity B = \(\frac{1}{2}\)
So, Answer will be C
Method 2
Coin is tossed 5 times => Total number of cases = \(2^5\) = 32
P(Getting more heads than tails) = (Cases in which we will get 3H + 4H + 5H) / 32
Cases in which we get 3H
Cases in which we get 3H can be found by putting 3 heads in any of the 5 slots _ _ _ _ _ _ _
This can be done in 5C3 ways = \(\frac{5!}{3!*(5-3)!}\) = \(\frac{5*4*3!}{3!*2!}\) = 10 ways
Cases in which we get 4H
Cases in which we get 4H is same as getting 1 Tail and this can be found by putting 1 tail in any of the 5 slots _ _ _ _ _ _ _
This can be done in 5 ways by putting T in any of the 5 slots => 5 ways
Cases in which we get 5H
This can be done in 1 way (HHHHH)
=> P(Getting more heads than tails) = \(\frac{10 + 5 + 1 }{ 32 }\) = \(\frac{16}{32}\) = \(\frac{1}{2}\)
=> Quantity A = Quantity B = \(\frac{1}{2}\)
So, Answer will be C
Hope it helps!
Watch the following video to learn How to Solve Probability with Coin Toss Problems
General Discussion
Re: Probability-A fair coin is flipped 5 times
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11 Aug 2015, 22:13
11 Aug 2015, 22:13
2
Expert Reply
goneewindy wrote:
Does anyone has a faster way of solving this problem instead of drawing out the tree?
A fair coin is flipped 5 times.
Quantity A: The probability of getting more heads than tails
Quantity B: 1/2
A fair coin is flipped 5 times.
Quantity A: The probability of getting more heads than tails
Quantity B: 1/2
There are 32 sample solutions in the solution set of the 5 coin toss. Since \(2^5 = 35\)
Now there are 5 coins so number of Heads can either be greater than or less than Tails. Since each head or tail is equally likely the probability of getting more heads is 0.5 or 16/32. The sample space is divided in two groups, "heads are more" and "tails are more".
If there had been even number of coins say 6. The sample space would have been 64. Now it wold have been divided into 3 groups, "heads are more" "tails are more" and "heads are equal to tails".
Now the "equal" case has 20 solutions. \(6!/3!*3!\) =20. The remaining 44 sample space points are equally likely to have heads or tails more.
Don't use logic unless you are well versed with probability theory. It is always easier and safer to draw the tree if you have the time.
