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Find the probability distribution of
(i) number of heads in two tosses of a coin
(ii) number of tails in the simultaneous tosses of three coins
(iii) number of heads in four tosses of a coin
Asked by Pethanan Last Modified
Answer (i) When one coin is tossed twice, the sample space is
{HH, HT, TH, TT}
Let X represent the number of heads.
∴ X (HH) = 2, X (HT) = 1, X (TH) = 1, X (TT) = 0
Therefore, X can take the value of 0, 1, or 2.
It is known that,
P (X = 0) = P (TT) 
P (X = 1) = P (HT) + P (TH)
P (X = 2) = P (HH)
Thus, the required probability distribution is as follows.
| X | 0 | 1 | 2 |
| P (X) |
|
|
|
(ii) When three coins are tossed simultaneously, the sample space is 
Let X represent the number of tails.
It can be seen that X can take the value of 0, 1, 2, or 3.
P (X = 0) = P (HHH) = 
P (X = 1) = P (HHT) + P (HTH) + P (THH) = 
P (X = 2) = P (HTT) + P (THT) + P (TTH) =
P (X = 3) = P (TTT) = 
Thus, the probability distribution is as follows.
| X | 0 | 1 | 2 | 3 |
| P (X) |
|
|
|
|
(iii) When a coin is tossed four times, the sample space is
Let X be the random variable, which represents the number of heads.
It can be seen that X can take the value of 0, 1, 2, 3, or 4.
P (X = 0) = P (TTTT) = 
P (X = 1) = P (TTTH) + P (TTHT) + P (THTT) + P (HTTT)
= 
P (X = 2) = P (HHTT) + P (THHT) + P (TTHH) + P (HTTH) + P (HTHT)
+ P (THTH)
= 
P (X = 3) = P (HHHT) + P (HHTH) + P (HTHH) P (THHH)
=
P (X = 4) = P (HHHH) = 
Thus, the probability distribution is as follows.
| X | 0 | 1 | 2 | 3 | 4 |
| P (X) |
|
|
|
|
|
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