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A.\[\dfrac{1}{4}\]

B.\[\dfrac{1}{2}\]

C.\[\dfrac{2}{3}\]

D.\[\dfrac{3}{4}\]

Answer
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In this question first find out all the possible outcomes when three coins will be tossed together and then find probability.

Given that a coin is tossed 3 times, so first create a sample space for all possible outcomes

\[S = \left\{ {HHH,HHT,HTH,HTT,THH,THT,TTH,TTT} \right\}\]

\[n\left( S \right) = 8\]

It means that there are a total of 8 different outcomes which are possible, where H represents the head and T represents Tail,

Outcome HHH means that when a coin each time head occurs whereas for HTH first head occurs next is tails and then again a head.

Now the events where 2 head and 2 tails occurred are

(2 Heads and 1 Tail) \[{S_{2H}} = \left\{ {HHT,HTH,THH} \right\} = 3\]

(2 Tails and 1 Head) \[{S_{2T}} = \left\{ {HTT,THT,TTH} \right\} = 3\]

So, the total sample space for 2 heads and 2 tails

\[{S_{2H}} + {S_{2T}} = 3 + 3 = 6\]

Hence the probability of getting 2 Heads and 2 Tails is\[ = \dfrac{6}{8} = \dfrac{3}{4}\].

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