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Find the probability distribution of the number of heads in two tosses of a coin.

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Last updated date: 24th Jun 2024
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Answer
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Hint: Firstly, we need to find the sample space for tossing of a coin, after that we can find the probability of getting heads for each event. This data can then be combined to form the required probability distribution table.
Formula to be used:
 $ P = \dfrac{f}{T} $ where, P is the probability, f and T are favorable and total outcomes respectively.

Complete step-by-step answer:
We are tossing two coins at a time, so the sample space i.e. possible outcomes every time are:
S = { HH, TT, TH, HT }
Where, H and T denotes heads and tails respectively.
If X denotes number of heads, then the value of X for each event can be given as:
X (HH) = 2 [2 heads in this event]
X (TH) = 1 [1 head in this event]
X (HT) = 1 [1 heads in this event]
X (TT) = 0 [0 heads in this event]
Thus, there are 3 possible values of X i.e. 0, 1, 2 and 4 total outcomes.
The total outcomes will be 4 and the favorable outcomes will be the occurrence of that value of X in the above data.
Then probability of each event using the formula $ P = \dfrac{f}{T} $ can be given as:
When X = 2,
 $ \Rightarrow P\left( X \right) = \dfrac{1}{4} $ as 2 appears only 1 time in the data.
When X = 1,
 $ \Rightarrow P\left( X \right) = \dfrac{2}{4} \to \dfrac{1}{2} $ as 1 appears 2 times in the data.
When X = 0,
 $ \Rightarrow P\left( X \right) = \dfrac{1}{4} $ as 0 appears only 1 time in the data.
Now, making the probability distribution using this information:

X012
P (X) $ \dfrac{1}{4} $ $ \dfrac{1}{2} $ $ \dfrac{1}{4} $

This is the required probability distribution of the number of heads in two tosses of a coin.

Note: We generally take variable X to denote the required outcome for calculation of probability distribution. We can also denote the probability of X along with its value like P (X = 1) for the value of X as 1.
We can also include the number of occurrences in the table, then the table would look like:

X012
Number of occurrences of heads121
P (X) $ \dfrac{1}{4} $ $ \dfrac{1}{2} $ $ \dfrac{1}{4} $