If you flip a coin 16 times, what is the probability that yo

Question

I remember encountering a problem like this somewhere, and I could not solve it. I don't have the problem with me right now, so no answer choices. But can anybody explain?

If you flip a coin 16 times, what is the probability that you will get 8 heads and 8 tails?

Is this even GMAT doable?

openbox1 · Answer

Probability of getting 8 heads in the first 8 throws and 8 tails in the following order of HHHHHHHHTTTTTTTT = 1/(2^8) x 1/(2^8) = 1 / (2^16)
Number of ways to rearrange HHHHHHHHTTTTTTTT
= 16!/(8! x 8!)

Simplify out 1/(2^16) x 16!/(8! x 8!)
= (3^2 x 5 x 11 x 13) / 2^15

Highly unlikely that they will give you such high numbers to work out.

fluke · Answer

Binomial distribution:
 C^{16}_{8}*\frac{1}{2^8}*\frac{1}{2^8}

manishgeorge · Answer

Fluke, could you explain ?

fluke · Answer

Frankly, I have memorized the formula and when/where to apply that. How does it give the exact count; for that you'd have to use a small set rather than these astronomical numbers. Let's just flip the coin 4 times and get exactly one head. C^4_1*\frac{1}{2^1}*\frac{1}{2^3}=\frac{4}{16}=\frac{1}{4} TTTT TTTH TTHT TTHH THTT THTH THHT THHH HTTT HTTH HTHT HTHH HHTT HHTH HHHT HHHH You see; only 4 of these have exactly 1 head. Let's see the components: Probability(Head)=\frac{1}{2} Probability(Not-Head/Tail)=1-\frac{1}{2}=\frac{1}{2} Number of times we expect the head to appear exactly: n Total trials : t So, Ways=C^t_n*Probability(Head)^n*Probability(Not-Head)^{(t-n)} In the original question: n=8 t=16 Ways=C^{16}_8*Probability(Head)^8*Probability(Not-Head)^{(16-8)} A better place to learn about Binomial Distribution is walker's tutorial on Probability: https://gmatclub.com/forum/math-probability-87244.html Topic: "Combination of independent and mutually exclusive events"

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If you flip a coin 16 times, what is the probability that yo

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