Question

Arushi, Mahi and Saina were fighting to get the first chance in a game. Arushi says, “Let us toss coins. If both heads appear, Mahi will take the first chance, if both tails appear, Saina will get it and if one head and one tail appear, I will get the chance.” Is her decision fair?

Answer 1

Here, we know that the total number of outcomes is 4 and the sample space is \left\\{ {{\text{HT,TH,HH,TT}}} \right\\}. We will use that the probability of the tossing of coins is given by dividing the number of outcomes of getting the favorable outcome by the total number of outcomes, that is; $P = \dfrac{{{\text{ Favorable outcomes}}}}{{{\text{Total number of outcomes}}}}$. Let us find the probability for $M$ getting 2 heads favoring Mahi, M = \left\\{ {{\text{HH}}} \right\\}, $S$ getting 2 heads favoring Saina, S = \left\\{ {{\text{TT}}} \right\\} and $A$ getting 1 head and 1 tail favoring Arushi, A = \left\\{ {{\text{TH,HT}}} \right\\} from the sample space. Then we will compare the probability of Mahi, Saina and Arushi to find that this method will be fair or not.

Complete step by step answer:

We are given that if both heads appear, Mahi will take the first chance, if both tails appear, Saina will get it and if one head and one tail appear, Arushi will get the chance.
When we toss two coins simultaneously then the possibility of outcomes are two heads, one head, and one tail, or two tails.
Since there the coin is tossed two times, so the total number of outcomes is 4.
We know the sample space of the experiment of tossing two coins is \left\\{ {{\text{HT,TH,HH,TT}}} \right\\}.
We know that the probability of the tossing of coins is given by dividing the number of outcomes of getting the favorable outcome by the total number of outcomes, that is; $P = \dfrac{{{\text{ Favorable outcomes}}}}{{{\text{Total number of outcomes}}}}$.
Let us assume that $M$ represents the event of getting 2 heads favoring Mahi and then we have M = \left\\{ {{\text{HH}}} \right\\} from the sample space.
We will now find the probability of getting two heads for Mahi using the above formula of probability.
$P\left( M \right) = \dfrac{1}{4}$
Let us assume that $S$ represents the event of getting 2 heads favoring Saina and then we have S = \left\\{ {{\text{TT}}} \right\\} from the sample space.
We will now find the probability of getting two heads for Saina using the above formula of probability.
$P\left( S \right) = \dfrac{1}{4}$
Let us now assume that $A$ represents the event of getting 1 head and 1 tail favoring Arushi and then we have A = \left\\{ {{\text{TH,HT}}} \right\\} from the sample space.
We will now find the probability of getting one head and one tail for Arushi using the above formula of probability.

$$\Rightarrow P\left( A \right) = \dfrac{2}{4} \\\ \Rightarrow P\left( A \right) = \dfrac{1}{2} \\\$$

After comparing the probability of Mahi, Saina, and Arushi, we found out that this method will result in more beneficial to Arushi.
Hence, Arushi’s decision was not fair as she kept two cases favorable to her and one each for the other two friends.

Note: We know that a sample space is the set of all possible outcomes of an experiment. One should know that the probability is simply how likely something is to happen. Since the two coins are tossed so when we get one tail and one head, it means that either of the coins will have a tail or head, so there are 2 favorable cases for Arushi and 1 each for Mahi and Saina which is unfair.