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A fair coin is tossed 3 times. The probability of obtaining two heads will be


1. 2/8
2. 3/8
3. 1/2
4. 1/8

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Correct Answer - Option 2 : 3/8

Concept:

If an event E can occur in n(E) = k ways out of n(s) = n equally likely ways, then, \(\rm P \left( E \right) = \frac{{n\left( E \right)}}{{n\left( s \right)}} = \frac{k}{n}\).

Calculation:

The set, S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}  is a sample space for the given experiment.

There are 8 elements in the sample space.

The possible outcomes is E = {HHT, HTH, THH} 

n(E) = 3

Probability of obtaining two heads = \(\rm P \left( E \right) = \frac{{n\left( E \right)}}{{n\left( s \right)}} = \frac 3 8\)

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