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A game consists of tossing a one rupee coin $3$ times and noting its outcome each time. Hanif wins if all the tosses give the same result, i.e. three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.

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The given question is related to probability. Try to recall the formulae related to probability of simultaneous occurrence of independent events.

Complete step-by-step answer:
We are given the case of a coin toss game. We know, during a coin toss there are only two possible outcomes, i.e. heads or tails. The probability of occurrence of head in a single toss is given as $P(H)=\dfrac{1}{2}$ and similarly, the probability of occurrence of a tail in a single toss is given as $P(T)=\dfrac{1}{2}$. In case of three tosses, the possible outcomes are as follows:
No heads: (T, T, T)
One head: (T, T, H), (T, H, T), (H, T, T)
Two heads: (H, H, T), (H, T, H), (T, H, H)
Three heads: (H, H, H)
Now, total number of possible outcomes is equal to $8$, { (H, H, H), (T, T, H), (T, H, T), (H, T, T), (H, H, T), (H, T, H), (T, H, H), (T, T, T) }. We are asked to find the probability that Hanif will lose the game. It is given that Hanif will lose the game if the outcomes of all the tosses are not the same, i.e. if the outcomes of all tosses are not heads or tails. The number of all such cases is $6$, { (T, T, H), (T, H, T), (H, T, T), (H, H, T), (H, T, H), (T, H, H) }.
Now, we know, probability $P=\dfrac{Number\,of\,favorable\,outcomes}{Number\,of\,total\,possible\,outcomes}$.
So, the probability that Hanif will lose the game is given as \[P=\dfrac{6}{8}=\dfrac{3}{4}\].

Note: While calculating probability, make sure to check every possible outcome. Generally, students miss one or two possible outcomes, because of which the value of probability changes and the wrong answer is obtained.
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