In a simultaneous throw of two coins the probability of getting at least one head is

\frac{1}{2}

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\frac{1}{3}

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\frac{2}{3}

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\frac{3}{4}

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Solution

The correct option is D $\frac{3}{4}$
Here S= ${H H, H T, T H, T T}$
Let E= event of getting at least on head = ${H T, T H, H H}$
$∴ P (E) = \frac{n (E)}{n (S)} = \frac{3}{4}$

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In a simultaneous throw of two coins the probability of getting at least one head is

The correct option is D 34Here S= {HH,HT,TH,TT}Let E= event of getting at least on head ={HT,TH,HH}∴P(E)=n(E)n(S)=34

The correct option is D $\frac{3}{4}$
Here S= ${H H, H T, T H, T T}$
Let E= event of getting at least on head = ${H T, T H, H H}$
$∴ P (E) = \frac{n (E)}{n (S)} = \frac{3}{4}$