Question

In a set of 10 coins, 2 coins are with heads on both sides. A coin is selected at random from this set and tossed five times. If all five times, the result was heads, find the probability that the selected coins had heads on both sides.
(a) $\dfrac{1}{6}$
(b) $\dfrac{8}{9}$
(c) $\dfrac{2}{3}$
(d) $\dfrac{2}{10}$

Answer 1

Hint : First of all, we will consider two events, in which in first event we will calculate the probability of coin selected randomly and it have head on both sides and in another event we will find the probability of coin selected randomly and it does not have heads on both sides. Then we will find the probability that the coin that is tossed 5 time have heads on both sides and second event in which the coin tossed 5 times does not have head on both the sides and then we will find the final answer by using the formula of conditional probability which can be given as, $P\left( \dfrac{{{E}_{1}}}{A} \right)=\dfrac{P\left( {{E}_{1}} \right)P\left( \dfrac{A}{{{E}_{1}}} \right)}{P\left( {{E}_{1}} \right)P\left( \dfrac{A}{{{E}_{1}}} \right)+P\left( {{E}_{2}} \right)P\left( \dfrac{A}{{{E}_{2}}} \right)}$. Thus, in this way we will find our answer.

Complete step by step solution :
In question we are given that in a set of 10 coins, 2 coins are with heads on both the sides. Now, a coin is selected at random and then it is tossed five times and the result of the toss was heads all five times. we are asked to find the probability of selected coins having heads on both the sides.
So, here we will two events, in one event we might select coin having head on both the sides and in another event, we might select a coin that does not have head on both the sides, so both these events can be given mathematically as,
${{E}_{1}}=$ Event of selecting a coin having heads on both sides.
${{E}_{2}}=$ Event of selecting a coin which does not have a head on both sides.
$A=$ Event in which we get heads all the five times.
Now, on considering the event 1, in which we select a coin having head on both sides, it can be given mathematically as,
$P\left( {{E}_{1}} \right)=\left( \dfrac{{}^{2}{{C}_{1}}}{{}^{10}{{C}_{1}}} \right)$
${}^{2}{{C}_{1}}$ as only one coin is selected in one time out of 2 coins having head on both the sides and there are total 10 coins so we have ${}^{10}{{C}_{1}}$. Now, on solving the problem using ${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!r!}$, we will get,
$P\left( {{E}_{1}} \right)=\left( \dfrac{{}^{2}{{C}_{1}}}{{}^{10}{{C}_{1}}} \right)=\dfrac{2}{10}$ ( $\because$as ${}^{n}{{C}_{1}}=n$) …………………..(i)
Now, on considering the event 2, in which we select a coin which does not have heads on both the sides, we will have 1 coin out of 8 coins and total 10 coins which can be seen mathematically as,
$P\left( {{E}_{2}} \right)=\left( \dfrac{{}^{8}{{C}_{1}}}{{}^{10}{{C}_{1}}} \right)$
$P\left( {{E}_{2}} \right)=\left( \dfrac{{}^{8}{{C}_{1}}}{{}^{10}{{C}_{1}}} \right)=\dfrac{8}{10}$ ( $\because$as ${}^{n}{{C}_{1}}=n$) ……………………(ii)
In each time the probability of coin getting both heads is only one and the probability of coin which does not have head on both sides is $\dfrac{1}{2}$, which can be written as,
$P\left( \dfrac{A}{{{E}_{1}}} \right)=1$
$P\left( \dfrac{A}{{{E}_{2}}} \right)=\dfrac{1}{2}$
Now, as the coin is tossed 5 times, the expression can be given as,
$P\left( \dfrac{A}{{{E}_{1}}} \right)={{\left( 1 \right)}^{5}}$ ……………………(iii)
$P\left( \dfrac{A}{{{E}_{2}}} \right)={{\left( \dfrac{1}{2} \right)}^{5}}$ ………………..(iv)
Now, the probability of an event that the selected coin has heads on both the sides is, $P\left( \dfrac{{{E}_{1}}}{A} \right)$, on applying the formula of conditional probability which can be given as,
$P\left( \dfrac{{{E}_{1}}}{A} \right)=\dfrac{P\left( {{E}_{1}} \right)P\left( \dfrac{A}{{{E}_{1}}} \right)}{P\left( {{E}_{1}} \right)P\left( \dfrac{A}{{{E}_{1}}} \right)+P\left( {{E}_{2}} \right)P\left( \dfrac{A}{{{E}_{2}}} \right)}$
Now, on substituting the values from expression (i), (ii), (iii) and (iv) we will get,
$\Rightarrow P\left( \dfrac{{{E}_{1}}}{A} \right)=\dfrac{\dfrac{2}{10}\times {{\left( 1 \right)}^{5}}}{\dfrac{2}{10}\times {{\left( 1 \right)}^{5}}+\dfrac{8}{10}\times {{\left( \dfrac{1}{2} \right)}^{5}}}$
$\Rightarrow P\left( \dfrac{{{E}_{1}}}{A} \right)=\dfrac{2}{2+\dfrac{8}{32}}=\dfrac{32\times 2}{\left( 32\times 2 \right)+8}$
$\Rightarrow P\left( \dfrac{{{E}_{1}}}{A} \right)=\dfrac{64}{64+8}=\dfrac{64}{72}=\dfrac{8}{9}$
Thus, the probability of selecting a coin at random and it having heads on both sides is $\dfrac{8}{9}$.
Hence, option (b) is the correct answer.

Note : Students might make mistakes in writing the mathematical form of the given probability. Let’s take here we are said that 2 coins out of 10 have heads on both the sides and then the coin is tossed and probability of the coin does not have head on both sides is $P\left( \dfrac{A}{{{E}_{2}}} \right)={{\left( \dfrac{1}{2} \right)}^{5}}$, but instead of that students might write it as, $P\left( \dfrac{{{E}_{2}}}{A} \right)={{\left( \dfrac{1}{2} \right)}^{5}}$, and when we substitute this value in formula of conditional probability which can be given as, $P\left( \dfrac{{{E}_{1}}}{A} \right)=\dfrac{P\left( {{E}_{1}} \right)P\left( \dfrac{A}{{{E}_{1}}} \right)}{P\left( {{E}_{1}} \right)P\left( \dfrac{A}{{{E}_{1}}} \right)+P\left( {{E}_{2}} \right)P\left( \dfrac{{{E}_{2}}}{A} \right)}$. So, we can see that it is completely opposite to our answer and wrong also, so students must be careful while using formulas and work properly.