havok wrote:If the probability of rain on any given day in City X is 50 percent, what is the probability that it rains on exactly 3 days in a 5-day period?

total no. of outcomes= 2^5, (as for each day they're two option either rain or no rain)

now we want 3 exact rainy days, let rainy day be represented by x and non rain days be represented by y,
therefore we need to arrange 5 objects of which 3 are of one type and 2 are of others, hence required arrangements would be 5!/2!*3!=10;

therefore required probability is 10/2^5=5/16..!!!

Hi Havok here it goes

The probability that it rains exactly 3 times in 5 days.
R R R D D (rain rain rain dry dry)
is (1/2)^5 = 1/32 ( prob of raining 1/2, dry 1/2
but R R R D D can also be arranged ( they might not be in sequence)

There are 5 C 3 = 5 C 2 ways of doing that,
hence 10

So the answer is 10 * 1/32 (same probability for each
of the ten) which equals 10/32
which reduces to 5/16

this problem can be done considering a coin HEADS AND TAILS.


Let me know if you need any further help on this.

For a discussion of coin flip and pseudo-coin flip questions (the question on this thread is one of the latter), check out:

https://www.beatthegmat.com/coin-flip-qu ... html#75414