Question
In a simultaneous throw of two coins, the probability of getting at least one head is-
Answer: Option D
Here S = {HH, HT, TH, TT}
Let E = event of getting at least one head = {HT, TH, HH}
$$\therefore P\left( E \right) = \frac{{n(E)}}{{n(S)}} = \frac{3}{4}$$
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Here S = {HH, HT, TH, TT}
Let E = event of getting at least one head = {HT, TH, HH}
$$\therefore P\left( E \right) = \frac{{n(E)}}{{n(S)}} = \frac{3}{4}$$
Was this answer helpful ?
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