Two unbiased coins are tossed simultaneously. The probability of getting two heads

\frac{1}{2}

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\frac{1}{4}

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\frac{3}{4}

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None

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Solution

The correct option is B $\frac{1}{4}$
Favourable number of outcomes, getting two heads = $1$ i.e {HH}
Total number of outcomes = $6$ i.e {HH, HT, TH, TT}
Probability $= \frac{1}{4}$

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