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probability coin question

Expert replies
by fangtray » Wed Sep 21, 2011 4:05 am
A coin is tossed five times. What is the probability of obtaining exactly two heads in the five tosses?

thanks.
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Source: — Problem Solving |

by mankey » Wed Sep 21, 2011 4:24 am
Is the answer equal to 5/2?

5C2 * (1/2)*(1/2)=5/2

What is the OA?

Thanks
Mankey
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by fangtray » Wed Sep 21, 2011 4:30 am
mankey wrote:Is the answer equal to 5/2?

5C2 * (1/2)*(1/2)=5/2

What is the OA?

Thanks
Mankey
this wasnt a multiple choice question i just wanted to know how to do it..i dont know if the probability is 5/2...a 250% chance that will happen? maybe i'm confused here..
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by knight247 » Wed Sep 21, 2011 4:32 am
When a coin is tossed FIVE times there are exactly 2^5=32 possible outcomes

Out of which we need EXACTLY two heads. So we need HHTTT or the different permutations of this sequence. We permute it as 5!/(2!*3!)=5*4/2=10 desired outcomes

10/32=5/16
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by GMATGuruNY » Wed Sep 21, 2011 4:45 am
fangtray wrote:A coin is tossed five times. What is the probability of obtaining exactly two heads in the five tosses?

thanks.
P(exactly n times) = P(one way) * total possible ways.

P(one way):
One way to get exactly 2 heads is to get heads on the first 2 flips and tails on the last 3 flips.
P(HHTTT) = 1/2 * 1/2 * 1/2 * 1/2 * 1/2 = 1/32.

Total possible ways:
Any arrangement of the letters HHTTT will yield exactly 2 heads.
Thus, to account for all the ways to get exactly 2 heads, the result above needs to be multiplied by the number of ways to arrange HHTTT.
Number of ways to arrange HHTTT = 5!/(2!3!) = 10.

Multiplying the results above, we get:
P(exactly 2 heads) = 10 * 1/32 = 10/32 = 5/16.
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by fangtray » Wed Sep 21, 2011 4:51 am
for HHTTT are you guys using 5!/2!(5-2)! or 5!/3!(5-3)!

i wanna know in case the numbers are different on a diff quesiton
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by GMATGuruNY » Wed Sep 21, 2011 4:57 am
fangtray wrote:for HHTTT are you guys using 5!/2!(5-2)! or 5!/3!(5-3)!

i wanna know in case the numbers are different on a diff quesiton
When there are identical elements in an arrangement, we have to divide by the number of ways to arrange the identical elements.

Number of ways to arrange HHHHH = 5!/5! = 1.
Number of ways to arrange HHHHT = 5!/4! = 5.
Number of ways to arrange HHHTT = 5!/3!2! = 10.
Number of ways to arrange HHTTT = 5!/2!3! = 10.
Number of ways to arrange HTTTT = 5!/4! = 5.
Number of ways to arrange TTTTT = 5!/5! = 1.
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My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
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