Question

There are 3 fair coins and 1 false coin with tails on both sides. A coin is chosen at random and tossed 4 times. If ‘tail’ occurs in all 4 times, then the probability that the false coin has been chosen for tossing is.

Answer 1

Hint: At first we find the probability of choosing a false and fair coin then we’ll find the probability coming 4 tails given that coin has already been chosen.
Now, we’ll use Bayes theorem to get the answer which states that \[{E_1},{\text{ }}{E_2}\;, \ldots ,{\text{ }}{E_n}\] be non empty sets of sample space S such that \[{E_1} + {\text{ }}{E_2}\; + \ldots {\text{ + }}{E_n} = S\]
Then the conditional probability $P({E_1}|A) = \dfrac{{P({E_1})P(A|{E_1})}}{{P({E_1})P(A|{E_1}) + P({E_2})P(A|{E_2}).......... + P({E_n})P(A|{E_n})}}$ where A be any event.

Complete step by step Answer:

Given data: There are 3 fair coins and 1 false coin with tails on both sides
We know that the probability of any event$ = \dfrac{{events(favourable)}}{{events(total)}}$
Therefore the probability of choosing the false coin$[P({A_1})]$$ = \dfrac{1}{4}$
And the probability of choosing a fair coin$[P({A_2})]$$ = \dfrac{3}{4}$
Probability of any event L given that M has already happened is given by$ = P(L|M)$
Now, let A be the event that 4 tails occur after choosing a coin.
The probability that all 4 tails occur when the false coin is chosen$ = P(A|{A_1})$
$ = 1 \times 1 \times 1 \times 1.............................(i)$
\[ = 1\]
Therefore the probability that all 4 tails occur when the fair coin is chosen$ = P(A|{A_2})$
$ = \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2}........................(ii)$
\[ = \dfrac{1}{{16}}\]
Now, using Bayes theorem
Probability that false coin is chosen given 4 tails occur on tossing$P({A_1}|A) =
\dfrac{{P({A_1})P(A|{A_1})}}{{P({A_1})P(A|{A_1}) + P({A_2})P(A|{A_2})}}$
$ = \dfrac{{\dfrac{1}{4}(1)}}{{\dfrac{1}{4}(1) + \dfrac{3}{4}\left( {\dfrac{1}{{16}}} \right)}}$
Multiplying numerator and denominator by 64
$ = \dfrac{{16}}{{16 + 3}}$
$ = \dfrac{{16}}{{19}}$
Therefore, the required probability\[ = \dfrac{{16}}{{19}}\]

Note: Remember for each individual case we multiply the probability as we’ve done in equation(i) and (ii), do not do the addition here as the probability for the further tails to come is dependent on that first tail as if the first tail comes on the toss then only the second tail can be there in second or on further tosses, so remember always where we multiply or add the respective probabilities.