Binomial trials and binomial distribution:
Consider a random experiment the outcomes of which can be classified as success or failure. It means that the experiment results in 2 outcomes only E1(success) or E2 (failure). Assume that experiment can be repeated many times, probability of success or failure in any trial are p and q (p + q = 1) and do not vary from trial to trial and finally different trials are independent. This type of experiment is called as Binomial experiment and trials are said to be binomial trials. For example tossing of a fair coin many times, each time outcome would be either a success or failure.
A probability distribution representing the binomial trials is said to binomial distribution.
Let us consider a Binomial experiment which is repeated 'n' times. Let probability of success and failure in any trial be p and q respectively and we are interested in probability of occurrence of 'r' successes in these n trials. Now number of ways of choosing 'r' success in 'n' trials = nCr. The probability of 'r' successes and (n-r) failures is pr×qn-r. Therefore probability of having exactly r successes = nCr×pr×qn-r
Let 'X' be random variable representing number of successes, then
P(X = r) = nCr×pr×qn-r (r = 0, 1, 2, L , n)
1 = (p + q)n = nC0p0qn + nC1p1qn -1 + nC2p2qn -2 + ....... + nCrprqn -r + ....... + nCnpn X -> Number of success0, 1, 2,......., r,........... n
Remarks:
- The probability of utmost 'r' successes in n trials =
- The probability of atleast 'r' successes in n trials =
- The probability of having 1st success at rth trial = p×qr-1
Example: A die is thrown 7 times. What is chance that the odd number turns up (i) exactly 4 times (ii) at least 4 times.
Sol: The probability of success =
(i) For the exactly 4 successes required probability =
(ii) For atleast 4 successes the required probability
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