Question

a coin tossed 25 times and data is recorded as follows, Heads:15 tails:10 find the probability of not getting head

Answer 1

If you are tossing a fair coin, then this is given by:

(1/2)14.(1/2)6=(1/2)20(1/2)14.(1/2)6=(1/2)20

as each toss is statistically independent of the others.

But this gives us information about one particular trial of the experiment. What is the probability that you would get 14 heads in general(i.e for the process as a whole, as opposed to a particular realization)?

For this you would have to consider the number of ways in which the 14 heads may occur. Now you might think that this is given by20P1420P14. But this is not correct. Why?

Because we can't distinguish between consecutive heads or tails. For example in the sequence given above,

HHHHHHHHHHHHHHTTTTTT

one could permute the heads and tails among themselves, but the sequences wouldn't be distinguishable. This is true for any number of consecutive heads. For example, consider the following sequence of 14 heads:

HHHTTTTHHHTTHHHHHHHH

You could again permute the heads and tails among themselves, but the sequences wouldn't be distinguishable. So the number of ways in which 14 heads may occur is not 20P1420P14 but 20C1420C14.

So the probability of getting 14 heads in any trial of the experiment is given by:

20C14.(1/2)14.(1/2)6=20C14.(1/2)2020C14.(1/2)14.(1/2)6=20C14.(1/2)20

This is however the case for a fair coin. Suppose you are using a biased coin in which the probability of getting heads is more than that of getting tails(say 3/4 for heads and 1/4 for tails), then the probability of getting 14 heads would be:

20C14.(3/4)14.(1/4)620C14.(3/4)14.(1/4)6

In general for a coin toss experiment having n trials, the probability of getting kheads is given by:

nCk.pk.(1−p)n−knCk.pk.(1−p)n−k

wherepp is the probability of getting heads in a single coin toss.

To calculate the probability of getting more than 14 heads, you would have to calculate the cumulative probability of getting 15, 16, 17, 18, 19 and 20 heads. One could calculate these probabilities in the way explained above and then sum them up.
hope this helps you

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Answer 2

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