Question

A coin is tossed three times. Find $P\left( \dfrac{A}{B} \right)$ in the given cases:
A = At least two heads, B = At most two heads.

Answer 1

Hint: In order to solve this question, we need to know that if a coin is tossed for n number of times, then the total possible outcomes will be ${{2}^{n}}$. Also, we need to remember a few formulas of probability like, $\text{Probability=}\dfrac{\text{Favourable outcomes}}{\text{Total outcomes}}$, $P\left( \dfrac{A}{B} \right)=\dfrac{P\left( A\cap B \right)}{P\left( B \right)}$. By using these concepts, we can find the answer to this question.

Complete step-by-step solution -
In this question, we have been asked to find the value of $P\left( \dfrac{A}{B} \right)$ for the event of tossing a coin 3 times, where A = at least two heads, B = at most two heads. Now, we know that for tossing a coin n times, the possible number of outcomes are ${{2}^{n}}$. So, for tossing a coin 3 times, the total number of outcomes will be ${{2}^{3}}=8$. And all the possible outcomes are HHH, HHT, HTH, THH, HTT, THT, TTH and TTT.
Now, we have been given 2 cases, A = at least 2 heads and B = at most 2 heads. So, we can say that the outcomes for A are {HHH, HHT, HTH, THH} and the outcomes for B are {TTT, TTH, THT, HTT, HHT, HTH, THH}. So, we can say that the outcomes for $A\cap B$ are {HHT, HTH, THH}. Now, we know that the probability of an event is given by, $P\text{=}\dfrac{\text{favourable outcomes}}{\text{total outcomes}}$. We know that for $A\cap B$ and B as the events, total outcomes are 8. So, we can say,
$P\left( A\cap B \right)=\dfrac{3}{8}$ and $P\left( B \right)=\dfrac{7}{8}$
Now, we know that $P\left( \dfrac{A}{B} \right)$ is calculated by $\dfrac{P\left( A\cap B \right)}{P\left( B \right)}$. So, we can say,
$P\left( \dfrac{A}{B} \right)=\dfrac{\dfrac{3}{8}}{\dfrac{7}{8}}=\dfrac{3\times 8}{7\times 8}=\dfrac{3}{7}$
Hence, we can say that $P\left( \dfrac{A}{B} \right)=\dfrac{3}{7}$ for an event of tossing a coin 3 times and A = at least 2 heads and B = at most 2 heads.

Note: We can also solve this question by the definition of $P\left( \dfrac{A}{B} \right)$ which means the probability of occurring of A when B has occurred. So, if we know the possible cases of occurrence of B, then we will find the number of cases which satisfies the condition of A then we will divide the number of cases of B satisfying A by the number of possible outcomes for B.