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# Three different coins are tossed together. Find the probability of getting:A.Exactly two headsB.At least two headsC.At least two tails  Verified
Hint: Probability is nothing but a ratio which can be written as $\dfrac{{Number{\text{ of favourable cases}}}}{{Number{\text{ of total cases}}}}$. Count the total number of cases. on each coin we can get either head or tail. Which is 2 in count. Think how we can use this information.
As the given scenario, the total possible cases are HHH, THH, HTH, HHT, TTH, THT, HTT, TTT which is 8 in total. We know that probability is nothing but a ratio. Formula to find the probability is $\dfrac{{Number{\text{ of favourable cases}}}}{{Number{\text{ of total cases}}}}$. We’ll use this formula as follows:
a) In the first case, exactly two heads are in our favour. It means favourable cases are THH, HTH, HHT which is 3 in total. Hence the probability of getting exactly two heads is $\dfrac{{Number{\text{ of favourable cases}}}}{{Number{\text{ of total cases}}}} = \dfrac{3}{8}$.
b) In the second case at least two heads in our favour. What does at least two heads mean? It means, number of heads could be more than two. Hence the favourable cases are THH, HTH, HHT, HHH which is 4 in total. Hence the probability of getting at least two heads is $\dfrac{{Number{\text{ of favourable cases}}}}{{Number{\text{ of total cases}}}} = \dfrac{4}{8} = \dfrac{1}{2}$
c) In the third case, at least two tails in our favour. What does at least two tails mean? It means, number of tails could be more than two. Hence the favourable cases are TTH, THT, HTT, TTT which is 4 in total. Hence the probability of getting at least two tails is $\dfrac{{Number{\text{ of favourable cases}}}}{{Number{\text{ of total cases}}}} = \dfrac{4}{8} = \dfrac{1}{2}$