Question

A biased coin which comes up heads three times as often as tails is tossed. If it shows heads, a chip is drawn from urn-I which contains 2 white chips and 5 red chips. If the coin comes up the tail, a chip is drawn from the urn-II which contains 7 white and 4 red chips. Given that a red chip was drawn, what is the probability that the coin came up heads?

Answer 1

Hint: To solve this question, we will use Bayes theorem whose formula is$P(\dfrac{{{E_1}}}{A}) = \dfrac{{P(\dfrac{A}{{{E_1}}}).P({E_1})}}{{P(\dfrac{A}{{{E_1}}}).P({E_1}) + P(\dfrac{A}{{{E_2}}}).P({E_2})}}$. We will also use conditional probability which is used to find the probability of an event when the other event is completed. Formula of conditional probability is $P(\dfrac{A}{B})$,where B is the event that has happened.

Complete step by step solution:
Now, let probability that the heads come up = p
Also, let probability that tails come up = q.
According to the question,
Probability of heads = 3 x probability of tails.
Therefore, p = 3q. … (1)
Now, Probability of tails come up = Probability when heads not come = 1 – p
Therefore, we get q = 1 – p
Putting this value of q in the equation (1), we get
p = 3(1 – p)
4p = 3
p = $\dfrac{3}{4}$ and q = $\dfrac{1}{4}$
Now, when a coin is tossed a chip is drawn from the urn. So, let A be the event when a chip is drawn from the urn.
When heads come up, a chip is drawn from urn-I. Now, to find the probability that a red chip is drawn from urn-I we will use conditional probability. Conditional probability is used to find the probability of an event when another event has completed. Now, urn-I has 2 white chips and 5 red chips.
So, probability of drawing a red chip when heads come up = P ($\dfrac{A}{p}$) = $\dfrac{5}{7}$
Similarly, when tails come up, a chip is drawn from urn-II. Urn-II contains 7 white chips and 4 red chips. So, probability of drawing a red chip when tails come up = P ($\dfrac{A}{q}$) = $\dfrac{4}{{11}}$
Now, we have to find the probability when a red chip is drawn and the coin comes up heads. So, to find this probability we will use Bayes Theorem. The formula of using Bayes theorem is,
$P(\dfrac{p}{A}) = \dfrac{{P(\dfrac{A}{p}).p}}{{P(\dfrac{A}{p}).p + P(\dfrac{A}{q}).q}}$
Applying values in the above formula, we get
$P(\dfrac{p}{A}) = \dfrac{{\dfrac{5}{7}.\dfrac{3}{4}}}{{\dfrac{5}{7}.\dfrac{3}{4} + \dfrac{4}{{11}}.\dfrac{1}{4}}}$ = $\dfrac{{15 \times 11}}{{15 \times 11 + 7 \times 4}}$ = $\dfrac{{165}}{{193}}$
So, probability of drawing a red chip and the coin came up heads = $\dfrac{{165}}{{193}}$

Note: Whenever we come up with such types of questions, we will use Conditional probability and Bayes theorem to solve the given problem. We will follow a few steps to solve such problems. First, we will find the probability of heads and tails coming up when a coin is tossed. The probability is not equal to each other because the coin is a biased coin. We will use the condition given in the question to find the probability. After finding probability, we will find the probability of drawing a red chip from the urn. After it, we will use Bayes theorem to find the required probability.