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If Jesse flips a coin seven times in a row, probablity that

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If Jesse flips a coin seven times in a row, probablity that

by gmattesttaker2 » Sun Jun 15, 2014 4:29 pm
Hello,

Can you please help with this:

If Jesse flips a coin seven times in a row, what is the probability that the result will be heads at least five times?

OA: [spoiler]29/128[/spoiler]

When I saw the word at-least 5 times, I used the following approach:

P(Heads at-least 5 times) = 1 - P(No heads at least 5 times)
= 1 - (1/2.1/2.1/2.1/2.1/2)
= 1 - (1/2)^5
= 1 - 1/32
= 31/32

However, it doesn't seem correct. Can you please assist?

Thanks a lot,
Sri
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by kvcpk » Sun Jun 15, 2014 5:57 pm
gmattesttaker2 wrote:Hello,

Can you please help with this:

If Jesse flips a coin seven times in a row, what is the probability that the result will be heads at least five times?

OA: [spoiler]29/128[/spoiler]

When I saw the word at-least 5 times, I used the following approach:

P(Heads at-least 5 times) = 1 - P(No heads at least 5 times)
= 1 - (1/2.1/2.1/2.1/2.1/2)
= 1 - (1/2)^5
= 1 - 1/32
= 31/32

However, it doesn't seem correct. Can you please assist?

Thanks a lot,
Sri
Hi Sri,

Here is an approach you could follow:
We are looking for a head atleast 5 times.
P(Atleast 5 times head) = P(Exactly 5 times head)+P(Exactly 6 times head) + P(Exactly 7 times head)

Let us calculate P(Exactly 5 times head) :
Probability of Getting 5 heads = 1/2 * 1/2* 1/2 * 1/2 * 1/2 = 1/2^5
Probability of Getting 2 tails = 1/2 * 1/2 = 1/4
Out of the 7 occurrences, We need only 5 occurrences of head. Hence we can obtain them in 7C5 ways.
Therefore, probability = 7C5 * 1/2^5 * 1/4 = 21/128

Applying the same logic Let us calculate P(Exactly 6 times head) :
Probability of Getting 6 heads = 1/2 * 1/2* 1/2 * 1/2 * 1/2 * 1/2= 1/2^6
Probability of Getting 1 tails = 1/2
Out of the 7 occurrences, We need only 6 occurrences of head. Hence we can obtain them in 7C6 ways.
Therefore, probability = 7C6 * 1/2^6 * 1/2 = 7/128

P(Exactly 7 times head) = 1/2 * 1/2 * 1/2 * 1/2 * 1/2 * 1/2 * 1/2 = 1/2^7

Adding all these, we get 21/128 + 7/128 + 1/128 = [spoiler]29/128[/spoiler]

Hope this helps!!
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by gmattesttaker2 » Tue Jun 17, 2014 11:24 pm
kvcpk wrote:
gmattesttaker2 wrote:Hello,

Can you please help with this:

If Jesse flips a coin seven times in a row, what is the probability that the result will be heads at least five times?

OA: [spoiler]29/128[/spoiler]

When I saw the word at-least 5 times, I used the following approach:

P(Heads at-least 5 times) = 1 - P(No heads at least 5 times)
= 1 - (1/2.1/2.1/2.1/2.1/2)
= 1 - (1/2)^5
= 1 - 1/32
= 31/32

However, it doesn't seem correct. Can you please assist?

Thanks a lot,
Sri
Hi Sri,

Here is an approach you could follow:
We are looking for a head atleast 5 times.
P(Atleast 5 times head) = P(Exactly 5 times head)+P(Exactly 6 times head) + P(Exactly 7 times head)

Let us calculate P(Exactly 5 times head) :
Probability of Getting 5 heads = 1/2 * 1/2* 1/2 * 1/2 * 1/2 = 1/2^5
Probability of Getting 2 tails = 1/2 * 1/2 = 1/4
Out of the 7 occurrences, We need only 5 occurrences of head. Hence we can obtain them in 7C5 ways.
Therefore, probability = 7C5 * 1/2^5 * 1/4 = 21/128

Applying the same logic Let us calculate P(Exactly 6 times head) :
Probability of Getting 6 heads = 1/2 * 1/2* 1/2 * 1/2 * 1/2 * 1/2= 1/2^6
Probability of Getting 1 tails = 1/2
Out of the 7 occurrences, We need only 6 occurrences of head. Hence we can obtain them in 7C6 ways.
Therefore, probability = 7C6 * 1/2^6 * 1/2 = 7/128

P(Exactly 7 times head) = 1/2 * 1/2 * 1/2 * 1/2 * 1/2 * 1/2 * 1/2 = 1/2^7

Adding all these, we get 21/128 + 7/128 + 1/128 = [spoiler]29/128[/spoiler]

Hope this helps!!
Hello KVCP,

Thank you very much for your detailed explanation.

Best Regards,
Sri
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by GMATinsight » Wed Jun 18, 2014 3:55 am
Probability = Favorable outcomes / Total outcomes

With 7 coins tossed, Total Outcomes = 2^7 = 128

Favorable out comes are as follows

1) Exactly 5 heads out of 7 coins = 7C5 = 21
2) Exactly 6 heads out of 7 coins = 7C6 = 7
3) Exactly 7 heads out of 7 coins = 7C7 = 1

Probability = (21+7+1)/128 = 29/128 ANSWER
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by Jeff@TargetTestPrep » Mon Apr 23, 2018 4:45 pm
gmattesttaker2 wrote:Hello,

Can you please help with this:

If Jesse flips a coin seven times in a row, what is the probability that the result will be heads at least five times?
We have 3 scenarios:

Scenario 1: 5 heads and 2 tails (e.g., H-H-H-H-H-T-T)

P(H-H-H-H-H-T-T) = (1/2)^7 = 1/128

Number of ways to arrange H-H-H-H-H-T-T = 7C5 = 7!/(5! x 2!) = (7 x 6)/2 = 21

Thus, the probability of this scenario is 1/128 x 21 = 21/128.

Scenario 2: 6 heads and 1 tail (e.g., H-H-H-H-H-H-T)

P(H-H-H-H-H-H-T) = (1/2)^7 = 1/128

Number of ways to arrange H-H-H-H-H-H-T = 7C6 = 7!/6! = 7

Thus the probability of this scenario is 1/128 x 7 = 7/128.

Scenario 3: all 7 heads (i.e., H-H-H-H-H-H-H)

P(H-H-H-H-H-H-H) = (1/2)^7 = 1/128

So, P(at least 5 heads) = 21/128 + 7/128 + 1/128 = [spoiler]29/128[/spoiler]

Jeffrey Miller
Head of GMAT Instruction
[email protected]

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