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

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gmattesttaker2: Legendary Member; Posts: 641; Joined: Tue Feb 14, 2012 3:52 pmFollowed by: 8 members; Thanked: 11 times

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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Source: Beat The GMAT — Problem Solving | Sun Jun 15, 2014 4:29 pm

kvcpk: Legendary Member; Posts: 1893; Joined: Sun May 30, 2010 11:48 pmFollowed by: 7 members; Thanked: 215 times

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!!

"Once you start working on something,
don't be afraid of failure and don't abandon it.
People who work sincerely are the happiest."
Chanakya quotes (Indian politician, strategist and writer, 350 BC-275BC)

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gmattesttaker2: Legendary Member; Posts: 641; Joined: Tue Feb 14, 2012 3:52 pmFollowed by: 8 members; Thanked: 11 times

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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GMATinsight: Legendary Member; Posts: 1100; Joined: Sat May 10, 2014 11:34 pmFollowed by: 24 members; Location: New Delhi, India; Thanked: 205 times

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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Jeff@TargetTestPrep: GMAT Instructor; Posts: 1462; Joined: Thu Apr 09, 2015 9:34 amFollowed by: 22 members; Location: New York, NY; Thanked: 39 times

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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