Question

Find probability distribution of no of heads in two tosses

Answer 1

hey!

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Let a coin is tossed n times.

• If we consider X to be the random variable representing the number of times head occurs then X follows a binomial distribution.

Let p be the probability of getting a head then,

• P(X=k) = (n C k) (p^k)(1-p)^(n-k)

• (n C k) = n/(n-k)*k

hope hhelp u

Answer 2

Explanation:

one coin is tossed twice, we can write the sample space as follows: S = {HH, HT, TH, TT}.

Since all these 4 events are equally likely, P(HH) = P(HT) = P(TH) = P(TT) = 1414

Let X be the number of heads in S such that →→ X(HH) = 2, X(HT) = 1, X(TH) = 1 and X(TT) = 0. It follows that:

P(X = 0) = P (TT) = 1414

P(X = 1) = P(TH, HT) = 24=1224=12

P(X = 2) = P(HH) = 1414

We can then construct the probability distribution table as follows:

XP(X)014112214