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

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

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.

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

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

\(= \frac{29}{128}\)
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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.
29/128 is the answer.

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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Solution as attached

Answer: Option B

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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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P = (5H + 6H + 7H)

5H Ways to choose*Choices: 7C5 * (1/2)^7 = 21/128
6H Ways to choose*Choices: 7C6 * (1/2)^7 = 7/128
7H Ways to choose*Choices: 7C7 * (1/2)^7 = 1/128

So 29/128, B.

Total number of possibilities = \(2^7 = 128\)

Number of ways to get 7 heads = 7C7 = 1

Number of ways to get 6 heads = 7C6 = 7

Number of ways to get 5 heads = 7C5 = 21

\(21 + 7 + 1 = 29\)

= \(\frac{29}{128}\)

Answer is B.
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