Question

A bag contains $n + 1$ coins. It is known that one of these coins shows heads on both sides, whereas the other coins are fair. One coin is selected at random and tossed. If the probability that toss results in heads is $\dfrac{7}{{12}}$​, then the value of $n$ is

Answer 1

Here the given question is based on the concept of probability. Given that there are two types of coins in the bag, we need to find the probability that the toss results in a head. For this, to find the probabilities of the two cases, that is the coin is from that $n$ coins or the coins is from that $n + 1$ coins and then find the probability that it will be heads and add both the probability cases.

Complete step by step answer:
Probability is a measure of the likelihood of an event to occur. Many events cannot be predicted with total certainty. We can predict only the chance of an event to occur i.e., how likely they are to happen, using it. Probability can range in from 0 to 1, where 0 means the event to be an impossible one and 1 indicates a certain event.
The probability formula is defined as the probability of an event to happen is equal to the ratio of the number of favourable outcomes and the total number of outcomes.
$\text{Probability of event to happen} P\left( E \right) = \dfrac{{\text{Number of favourable outcomes}}}{{\text{Total Number of outcomes}}}$
Consider the given question:
Given that out of coins there is one rigged coin and $n$ fair coins which implies the probability of getting a head on a rigged coin is 1.
The probability of getting a head on the fair coin is $\dfrac{1}{2}$ or $0.5$.
Given that the probability that the toss results in a head is $\dfrac{7}{{12}}$.
Hence, if one coin is drawn it can either be a rigged coin or a fair coin. Therefore, the required probability is
$\Rightarrow$ (Probability that the coin is drawn from one coin $\times$ probability of getting a head on rigged coin) $+$ (Probability that the coin is drawn from n coins $\times$ probability of getting a head on the fair coin) $= \dfrac{7}{{12}}$
$\Rightarrow \,\,\dfrac{1}{{n + 1}} \times 1 + \dfrac{n}{{n + 1}} \times \dfrac{1}{2} = \dfrac{7}{{12}}$
$\Rightarrow \,\,\dfrac{1}{{n + 1}} + \dfrac{n}{{2\left( {n + 1} \right)}} = \dfrac{7}{{12}}$
Multiply and divide by 2 to the first term of above equation, then we have
$\Rightarrow \,\,\dfrac{2}{{2\left( {n + 1} \right)}} + \dfrac{n}{{2\left( {n + 1} \right)}} = \dfrac{7}{{12}}$
$\Rightarrow \,\,\dfrac{{2 + n}}{{2\left( {n + 1} \right)}} = \dfrac{7}{{12}}$
Multiply 2 on both sides
$\Rightarrow \,\,\dfrac{{n + 2}}{{n + 1}} = \dfrac{7}{6}$
On cross multiplication, then we have
$\Rightarrow \,\,6\left( {n + 2} \right) = 7\left( {n + 1} \right)$
On multiplication, we have
$\Rightarrow \,\,6n + 12 = 7n + 7$
Subtract both side by 7
$\Rightarrow \,\,6n + 12 - 7 = 7n$
$\Rightarrow \,\,6n + 5 = 7n$
Or
$\Rightarrow \,\,7n = 6n + 5$
Subtract both side by $6n$
$\Rightarrow \,\,7n - 6n = 5$
On simplification, we get
$\therefore \,\,\,\,n = 5$
Hence, the required solution of $\,n = 5$.

Note:
Probability is a measure of the likelihood of an event to occur. Many events cannot be predicted with total certainty. We can predict only the chance of an event to occur i.e., how likely they are to happen, using it. Probability can range in from 0 to 1, where 0 means the event to be an impossible one and 1 indicates a certain event.