Question

Two coins are tossed simultaneously 50 times with the following frequencies of different outcomes:

Outcomes	2 heads	1 head	No heads
Frequency	13	26	11

Find the probability of each outcome. Show that the sum of the probabilities is equal to 1.

Answer 1

Hint: To solve this question we will use the probability of an even formula. Probability is given by \[\text{Probability}=\dfrac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}.\] Here, 50 times toss means the total number of outcomes is 50. Now, we will calculate each probability separately using the above-stated formula and add them to get the sum as 1.

Complete step-by-step solution:
We are given the table as

Outcomes	2 heads	1 head	No heads
Frequency	13	26	11

Two coins are tossed simultaneously 50 times. The total number of outcomes is 50. We have to find the probability of each outcome and for that, we will use the formula given below. The probability of an event is given by \[\text{Probability}=\dfrac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}.\]
Now, let \[{{P}_{1}}\] be the probability of having 2 heads, referring to the table we see that the number of favorable outcomes of \[{{P}_{1}}\] is 13. Using the above formula, we can write
\[\Rightarrow \text{Probability}\left( {{P}_{1}} \right)=\dfrac{13}{50}\]
\[\Rightarrow {{P}_{1}}=\dfrac{13}{50}.....\left( i \right)\]
Let \[{{P}_{2}}\] be the probability of having 1 head. Then observing the table, we see that the number of favorable outcomes \[{{P}_{2}}\] has 26.
\[\text{Probability}\left( {{P}_{2}} \right)=\dfrac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}\]
\[\Rightarrow {{P}_{2}}=\dfrac{26}{50}.....\left( ii \right)\]
Finally, let \[{{P}_{3}}\] be the probability of the event having no head. Referring to the table, we see that the number of favorable outcomes of \[{{P}_{2}}\] is 11. Using the above formula, we get,
\[\Rightarrow \text{Probability}\left( {{P}_{3}} \right)=\dfrac{11}{50}.....\left( iii \right)\]
Hence, we have obtained each probability separately using the probability formula. Finally, we have to show that the sum of the probability is equal to 1.
\[\Rightarrow \left( i \right)+\left( ii \right)+\left( iii \right)=1\]
So, adding equations (i), (ii), and (iii), we get,
\[\Rightarrow \dfrac{13}{50}+\dfrac{26}{50}+\dfrac{11}{50}=1\]
\[\Rightarrow \dfrac{50}{50}=1\]
Hence proved.

Note: The key point to note here in this question is that always the total probability of any event is 1. Even if the probability of 2 tails, 1 tail, and no tail was to be determined, then the sum would also be 1. Here, we had 2 coins tossed simultaneously, so only 3 possibilities that are 1 head, 2 heads, and 0 head are given as the sum to be 1.