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# If Jesse flips a coin seven times in a row, what is the probability th

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If Jesse flips a coin seven times in a row, what is the probability th  [#permalink]

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08 Jan 2016, 11:23
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If Jesse flips a coin seven times in a row, what is the probability that the result will be heads at least five times?

A. 21/128
B. 29/128
C. 35/128
D. 1/16
E. 1/4

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Re: If Jesse flips a coin seven times in a row, what is the probability th  [#permalink]

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08 Jan 2016, 21:13
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Question : Atleast 5 heads in 7 flip.

The total outcome of flip is = 2^7 = 128
For any Coins problem write the ask in the shown format.

HHHHHTT
HHHHHHT
HHHHHHH

Once you have written in the above mentioned format the answer is pretty straight.

HHHHHTT = [7!]/5!*2! = 21
HHHHHHT = [7!]/6! = 7
HHHHHHH = [7!]/[7!] = 1

Sum = 21+7+1 = 29/128
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Joined: 02 Aug 2009
Posts: 6547
If Jesse flips a coin seven times in a row, what is the probability th  [#permalink]

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08 Jan 2016, 20:33
4
If Jesse flips a coin seven times in a row, what is the probability that the result will be heads at least five times?

A. 21/128
B. 29/128
C. 35/128
D. 1/16
E. 1/4

Hi,
since the Q asks us atleast 5 times, we will have to find 5 times, 6 times and 7 times..
1) 5 times..
there are 7C5 ways that the heads can be placed in 7 times..
the prob of heads or tails is 1/2..
so prob in each way of head=$$(\frac{1}{2})^5*(\frac{1}{2})^2$$..
total ways=$$7C5*(\frac{1}{2})^5*(\frac{1}{2})^2= 21*(\frac{1}{2})^7$$..
2)6 times
$$7C6*(\frac{1}{2})^6*(\frac{1}{2})^1=7*(\frac{1}{2})^7$$.
3)7 times
$$7C7*(\frac{1}{2})^7*(\frac{1}{2})^0=1*(\frac{1}{2})^7$$...
Since the probability is 1/2, and only two case possible, here entire thing can be written as $$7C5+7C6+7C7=7*6/2+7+1=21+7+1=29$$ and total possibility = 2^7

total ways..
$$(21+7+1)*(\frac{1}{2})^7=\frac{29}{128}$$
B
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If Jesse flips a coin seven times in a row, what is the probability th  [#permalink]

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08 Jan 2016, 20:39
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If Jesse flips a coin seven times in a row, what is the probability that the result will be heads at least five times?

A. 21/128
B. 29/128
C. 35/128
D. 1/16
E. 1/4

Hi,
I have explained the way to get this formula, but you can use this ai almost all 'atleast times' probability..
This is also called binomial prob formula..
prob of doing a thing x times out of total n ways= $$nCx*p^x$$ *(1 − p)^n−x ,
where n denotes the number of trials and p denotes the success
probability.
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3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

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Re: If Jesse flips a coin seven times in a row, what is the probability th  [#permalink]

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03 May 2016, 02:06
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Total outcome is $$2^7=128$$

Desired outcomes: $$7C5 + 7C6 +7C7 = 21 + 7 + 1$$

$$= \frac{29}{128}$$
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Re: If Jesse flips a coin seven times in a row, what is the probability th  [#permalink]

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09 Oct 2016, 11:20
If Jesse flips a coin seven times in a row, what is the probability that the result will be heads at least five times?

A. 21/128
B. 29/128
C. 35/128
D. 1/16
E. 1/4

what a tricky damn one...
since order doesn't matter...i started with the number of ways we can get 5, 6, or even 7 heads...
5 heads = 7C5 = 21
6 heads = 7C6 = 7
7 heads = 7C7 = 1

total: 21+7+1 = 29.
this number, multiply by (1/2)^7.
1/2^7 = 1/128.
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Re: If Jesse flips a coin seven times in a row, what is the probability th  [#permalink]

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26 Oct 2017, 06:35
Probability = favorable outcomes / total outcomes

Favourable outcomes = no of times we get heads AT LEAST 5 times = no of times we get heads 5 times + no of times we get heads 6 times + no of times we get heads 7 times = 7C5 + 7C6 + 7C7 = 29

Total favorable outcomes = 2^7

So ans = 29/128

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Re: If Jesse flips a coin seven times in a row, what is the probability th  [#permalink]

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20 Apr 2018, 02:36
If Jesse flips a coin seven times in a row, what is the probability that the result will be heads at least five times?

A. 21/128
B. 29/128
C. 35/128
D. 1/16
E. 1/4

Solution as attached

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Re: If Jesse flips a coin seven times in a row, what is the probability th  [#permalink]

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23 Apr 2018, 17:37
If Jesse flips a coin seven times in a row, what is the probability that the result will be heads at least five times?

A. 21/128
B. 29/128
C. 35/128
D. 1/16
E. 1/4

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 = 29/128.

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Re: If Jesse flips a coin seven times in a row, what is the probability th &nbs [#permalink] 23 Apr 2018, 17:37
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