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shasadou
Joined: 12 Aug 2015
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Concentration: General Management, Operations
Schools: Johnson '18 (WL) Duke '18 (M) Tuck '18 (D)
GMAT 1: 640 Q40 V37
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WE:Management Consulting (Consulting)
08 Jan 2016, 11:23
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B
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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
A. 21/128
B. 29/128
C. 35/128
D. 1/16
E. 1/4
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
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
08 Jan 2016, 20:33
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shasadou
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
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
