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Binomial Distribution
Posted by topIITcoaching Experts
Last week we discussed Random Variable - mean and variance. Continuing on statistics, this week we are going to discuss Binomial Distribution.
To begin Binomial distribution, we must know that a lot of experiments are possible where there are only two outcomes (or we assume that any other outcome is ruled out). For example when we toss a coin the only possible outcomes are either a Head or a Tail (and we dont assume that the coin will fall vertically and remain upright, it will fall flat eventually). So, one of them can be termed as singlequotesuccesssinglequote and another as singlequotefailuresinglequote. Such trials which have only two outcomes are called Bernoulli Trial after Swiss scientist Jacob Bernoulli. In Bernoulli trial the outcome of each trial in independent of previous trials, the probability of success and failiue remains same in each trial.
For example, if we toss a coin then there are only two possible outcomes and the probability of a head or tail remains constant in each trial which is independent of the outcome of previous trial.
Binomial Distribution: Binomial distribution is a discrete probability distribution of a sequence of n consecutive experiments (or trials) where each trial has a chance of success of p and chance of failure is q (where p + q = 1).
For example we a toss a coin three times (here n = 3 ) and we want to know the probability of heads two times. In other words, if outcome HEAD is a success we are looking for two successes. The possible favorable outcomes will be HTH,HHT,THH i.e., chosing 2 places out of 3. It is possible in 3C2 ways.
So,
P(HTH) = p.q.p = p2q
P(HHT) = p.p.q = p2q
P(THH) = q.p.p = p2q
Total probability = p2q + p2q + p2q = 3p2q = 3C2p2q
P(X = 2) = 3C2p2q3-2
The above expression can be genralized as:
P(X = r) = 3Crprq3-r ( n = 3)
So, we get
If the number of trials is n the generalized formula becomes:
P(X = r) = nCrprqn-r
Example: Five dice are thrown simultaneoudsly. If the occurence of an even number in a single throw is considered a success, find the probability of at most 3 successes.
Solution: Probability of at most three success = probabilty of one success + probabilty of one successes + probabilty of three successes.
Here probabiliry of success, p = 3/6 = 1/2
Probability of failure,
= P(X=0) + P(X=1) + P(X=2) + P(X=3)
= 5C0p0q5 + 5C1p1q4 + 5C2p2q3 + 5C3p3q2
Mean and Variance of Binomial Distribution:
For a Binomial Variate X with success probability p and number of experiment n is given by:

Variance:

To begin Binomial distribution, we must know that a lot of experiments are possible where there are only two outcomes (or we assume that any other outcome is ruled out). For example when we toss a coin the only possible outcomes are either a Head or a Tail (and we dont assume that the coin will fall vertically and remain upright, it will fall flat eventually). So, one of them can be termed as singlequotesuccesssinglequote and another as singlequotefailuresinglequote. Such trials which have only two outcomes are called Bernoulli Trial after Swiss scientist Jacob Bernoulli. In Bernoulli trial the outcome of each trial in independent of previous trials, the probability of success and failiue remains same in each trial.
For example, if we toss a coin then there are only two possible outcomes and the probability of a head or tail remains constant in each trial which is independent of the outcome of previous trial.
Binomial Distribution: Binomial distribution is a discrete probability distribution of a sequence of n consecutive experiments (or trials) where each trial has a chance of success of p and chance of failure is q (where p + q = 1).
For example we a toss a coin three times (here n = 3 ) and we want to know the probability of heads two times. In other words, if outcome HEAD is a success we are looking for two successes. The possible favorable outcomes will be HTH,HHT,THH i.e., chosing 2 places out of 3. It is possible in 3C2 ways.
So,
P(HTH) = p.q.p = p2q
P(HHT) = p.p.q = p2q
P(THH) = q.p.p = p2q
Total probability = p2q + p2q + p2q = 3p2q = 3C2p2q
P(X = 2) = 3C2p2q3-2
The above expression can be genralized as:
P(X = r) = 3Crprq3-r ( n = 3)
So, we get
If the number of trials is n the generalized formula becomes:
P(X = r) = nCrprqn-r
Example: Five dice are thrown simultaneoudsly. If the occurence of an even number in a single throw is considered a success, find the probability of at most 3 successes.
Solution: Probability of at most three success = probabilty of one success + probabilty of one successes + probabilty of three successes.
Here probabiliry of success, p = 3/6 = 1/2
Probability of failure,
= P(X=0) + P(X=1) + P(X=2) + P(X=3)
= 5C0p0q5 + 5C1p1q4 + 5C2p2q3 + 5C3p3q2
Mean and Variance of Binomial Distribution:
For a Binomial Variate X with success probability p and number of experiment n is given by:

Variance:

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