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Coin Probabilty please help!

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Coin Probabilty please help!

by [email protected] » Thu Jul 31, 2008 6:53 am
Hi

I know this is probably a very easy question to answer but I always struggle with these types of questions can someone please explain a good strategy to approaching these types of question.


Five fair coins are tossed. What is the probability that exactly three of the coins land tails side up?

A)5/32
B)3/16
C)5/16
D)3/8
E)5/8
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by sudhir3127 » Thu Jul 31, 2008 6:59 am
My Answer is C.

here it goes.. exactly 3 tails...

in a flip of coin the probability of head/tail is 1/2

exactly 3 means

5C3*(1/2)3*(1/2)2....( i am using binomial theorem... )

10/32= 5/16,

do let me know if u have any doubts...
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by VP_RedSoxFan » Thu Jul 31, 2008 7:10 am
The standard approach for probability is to take the number of ways you can get your desired outcome divided by the total possible outcomes.

In this case, our desired "outcome" is exactly 3 heads. Since we don't care what order we get them in we can use the combination formula 5_C_3. This will give us the total number of ways to get our desired outcome of 3 heads. 5_C_3 = 10

Since each of the 5 coins can have 2 outcomes, the total number of possible arrangements of heads and tails for 5 coins is 2^5 = 32

Again, desired outcomes (10) divided by all possible outcomes (32) gives us a probability of 5/16.
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by dhanda.arun » Thu Jul 31, 2008 12:40 pm
The Answer is C
Total no of outcomes = 1/(2)^5= 1/32
Favorable outcomes = 5C3 = 10
so result is 10/32
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