Answer :

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 4th 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 pi and xi.


Let’s proceed to find mean and variance.


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


Variance is given by:


Variance = ∑ xi2pi – (∑xipi)2


first we need to find the products i.e. pixi and pixi2 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 =


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