A coin is tossed five times. What is the probability of obtaining exactly two heads in the five tosses?
thanks.
thanks.
Is the answer equal to 5/2?
5C2 * (1/2)*(1/2)=5/2
What is the OA?
Thanks
Mankey
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probability coin question
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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..mankey wrote:Is the answer equal to 5/2?
5C2 * (1/2)*(1/2)=5/2
What is the OA?
Thanks
Mankey
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
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
-
GMATGuruNY
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P(exactly n times) = P(one way) * total possible ways.fangtray wrote:A coin is tossed five times. What is the probability of obtaining exactly two heads in the five tosses?
thanks.
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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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
i wanna know in case the numbers are different on a diff quesiton
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GMATGuruNY
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When there are identical elements in an arrangement, we have to divide by the number of ways to arrange the identical elements.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
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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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
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.
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