Say, H represents event of getting a head and T represents getting a tail.

When we toss a coin 4 times we have the following possibilities:

{HHHH,HHHT,HHTH,THHH,HTHH,THHT,TTHH,HHTT,THTH…………,TTTT}

A total of 2⁴ = 16 possibilities.

∵ probability of getting a head or probability of getting a tail are independent events and P(GETTING A HEAD) = P(GETTING A TAIL) = 1/2

∴ P(Head in first toss) and P(Head in second toss) and P(head in third toss) and P(tail in 4^th toss) can be given by their individual products.

Note: P(AՈB) = P(A)P(B) where A and B are independent events.

As X is a random variable representing longest string of head occurring in 4 tosses.

∴ X can take following values:

X = 0 [ all tails (TTTT) ]

X = 1 [Longest string contains only 1 head e.g. (HTTT),(TTTH),(HTHT)..]

X = 2 [ Longest string contain only 2 head e.g. (HHTT),(HHTH),(THHT)…]

X = 3 [Longest string contain only 3 head e.g. ( HHHT) And (THHH)]

X = 4 [ Longest string contain 4 heads i.e. (HHHH) ]

Thus,

P(X=0) = 1/16

P(X=1) = 7/16 [by counting number of favourable outcomes as explained]

P(X=2) = 5/16

P(X=3) = 2/16

P(X=4) = 1/16

Now we have p_i and x_i.

Let’s proceed to find mean and variance.

Mean of any probability distribution is given by Mean = ∑x_ip_i

Variance is given by:

Variance = ∑ x_i²p_i – (∑x_ip_i)²

∴ first we need to find the products i.e. p_ix_i and p_ix_i² and add them to get mean and apply the above formula to get the variance.

Following table representing probability distribution gives the required products :

∴ Mean =

Variance =