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Binomial probability

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by crackgmat007 » Fri May 08, 2009 8:57 am
If a certain coin is flipped, the probability that the coin will land heads is 1/2. If the coin is flipped 5 times, what is the probability that it will land heads up on the first 3 flips and not on the
last 2 flips?
A. 3/5
B. 1/2
C. 1/5
D. 1/8
E. 1/32

I computed the probability of first 3 flips by using the below formula

5C3 * (1/2)^3 * (1/2)^2 =>This gives me probability of 3 flips being heads.

How to I go from here? Let me know. Tx.
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by DeepakR » Fri May 08, 2009 9:18 am
We know that P=Desired outcome/Total number of outcome

Here Total outcome = 2^5=32
Desired outcome = HHHTT = 1

Hence P = 1/32 = E.)

- Deepak
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Re: Binomial probability

by scoobydooby » Fri May 08, 2009 9:24 am
crackgmat007 wrote:If a certain coin is flipped, the probability that the coin will land heads is 1/2. If the coin is flipped 5 times, what is the probability that it will land heads up on the first 3 flips and not on the
last 2 flips?
A. 3/5
B. 1/2
C. 1/5
D. 1/8
E. 1/32

I computed the probability of first 3 flips by using the below formula

5C3 * (1/2)^3 * (1/2)^2 =>This gives me probability of 3 flips being heads.

How to I go from here? Let me know. Tx.

what you have done answers the question: what is the probability of getting exactly 3 heads in 5 flips?

but here, it is given that we want the first 3 flips to be heads and the other two to be tails=> there is only one way to get first 3 flips as heads,

so we do not need the combination 5C3
so it simply becomes 1*(prob of 3 heads)*(prob of 2 tails)
=>(1/2)^3*(1/2)2
=>(1/2)^5=1/32
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by Uri » Fri May 08, 2009 12:27 pm
the probability of getting head in the 1st toss=1/2
the probability of getting head in the 2nd toss=1/2
the probability of getting head in the 3rd toss=1/2
the probability of getting tail in the 4th toass=1/2
the probability of getting tail in the 5th toass=1/2
So, probability of getting heads in the first three tosses and tails in the last two tosses
=(1/2)^5=1/32
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