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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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

Total outcome is \(2^7=128\)

Desired outcomes: \(7C5 + 7C6 +7C7 = 21 + 7 + 1\)

\(= \frac{29}{128}\)
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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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Consistency is the Key

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.

Answer: B
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