Hard probability
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Hi,
# of ways of getting exactly 3 heads = 5C3 = 10
# of ways of getting exactly 4 heads = 5C4 = 5
# of ways of getting exactly 5 heads = 5C5 = 1
# of ways of getting at least 3 heads = 10 + 5 + 1 = 16
Total # of ways of getting outcomes = 2 power 5 = 32.
Hence, probability = 1/2.
Hope this helps. Thanks.
# of ways of getting exactly 3 heads = 5C3 = 10
# of ways of getting exactly 4 heads = 5C4 = 5
# of ways of getting exactly 5 heads = 5C5 = 1
# of ways of getting at least 3 heads = 10 + 5 + 1 = 16
Total # of ways of getting outcomes = 2 power 5 = 32.
Hence, probability = 1/2.
Hope this helps. Thanks.
Naveenan Ramachandran
4GMAT, Dadar(W) & Ghatkopar(W), Mumbai
4GMAT, Dadar(W) & Ghatkopar(W), Mumbai
- sumanr84
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General coin flip strategy,duongthang wrote:what is probability of getting at least 3 heads when flipping 5 coins
https://www.beatthegmat.com/coin-flip-qu ... 17911.html
- Patrick_GMATFix
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We can also arrive at the answer through reasoning. With 5 flips the following outcomes will have the same probability:
>> All heads = All tails
>> 1 tail & 4 heads = 1 head & 4 tails
>> 2 tails & 3 heads = 2 heads & 3 tails
Probability of at least 3 heads is the sum of all the probabilities on the left side (3 or 4 or all heads). This must equal the sum of probabilities on the right side. Since all outcomes must have a combined probability of 1, the sum of each side must be 1/2.
P (at least 3 heads) = 1/2
>> All heads = All tails
>> 1 tail & 4 heads = 1 head & 4 tails
>> 2 tails & 3 heads = 2 heads & 3 tails
Probability of at least 3 heads is the sum of all the probabilities on the left side (3 or 4 or all heads). This must equal the sum of probabilities on the right side. Since all outcomes must have a combined probability of 1, the sum of each side must be 1/2.
P (at least 3 heads) = 1/2
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