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# If a coin is tossed 3 times the probability of obtaining 2 heads or 2 tails is A.$\dfrac{1}{4}$B.$\dfrac{1}{2}$C.$\dfrac{2}{3}$D.$\dfrac{3}{4}$  Hint: A branch of mathematics which deals with numerical analysis of how likely an event is to occur, or the proportion is true. Probability of an event lies between 0 and 1, which means the chance of occurrence of an event increases as probability leads to 1 and if probability is 0 for an event then there is no chance of occurrence of that event.
In this question first find out all the possible outcomes when three coins will be tossed together and then find probability.

Complete step by step solution:
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$= \dfrac{6}{8} = \dfrac{3}{4}$.
Option (D) is the right answer

Note: The probability of all events in a sample space adds up to 1. Students should not get confused with the terms “OR” and “AND”, these are two different things associated with the probability. “OR” is associated with the summation of the probabilities whereas “AND” is associated with the product of the probability.

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