probability coin question

[fangtray]

A coin is tossed five times. What is the probability of obtaining exactly two heads in the five tosses?

thanks.

[mankey]

Is the answer equal to 5/2?

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

What is the OA?

Thanks
Mankey

[fangtray]

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..

[knight247]

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

[GMATGuruNY]

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.

[fangtray]

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

[GMATGuruNY]

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.