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Best answer
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13 votes

Ans D)

In 3 coins we are getting 2 heads and 1 tail (H,H,T),(H,T,H),(H,H,T)  =3! / 2! =3 ways

Now, probability =3/23 =3/8

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5 votes
5 votes

Three coins are tossed so number of outcomes=2*2*2=8

We are required to get probability if getting two heads and one tail.

Favourable possibilities = HHT , HTH , THH


Hence probability of getting two heads and one tail=$\frac{3}{8}$

Hence,Option(d)$\frac{3}{8}$ is the correct choice.

3 votes
3 votes
these are the possibilities hhh , hht , thh ,hth, tth, htt,tht, ttt

so required is 2 head 1 tail i.e- hht ,thh, hth

so probability is 3/8 option d
Answer:

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