Question

A man has $3$ coins A, B & C. A is fair coin. B is biased such that the probability of occurring head on it is $\dfrac{2}{3}$ C is also biased with the probability of occurring head as$\dfrac{1}{3}$. If one coin is selected and tossed three times, giving two heads and one tail, find the probability that the chosen coin was A.
\[
  A.\,\dfrac{9}{{25}} \\
  B.\;\dfrac{3}{5} \\
  C.\;\dfrac{{27}}{{125}} \\
  D.\;\dfrac{1}{3} \\
 \]

Answer 1

Hint:
Probability of any given event is equal to the ratio of the favourable outcomes with the total number of the outcomes. Probability is the state of being probable and the extent to which something is likely to happen in the particular favourable situations. Here, we will find the probability of each of the three coins in the favourable pattern of Head, Head and Tail.

Complete step by step solution:
Give that: Coin A is the fair coin.
Therefore, the probability of occurring head for coin A is $P{(H)_A} = \dfrac{1}{2}$
(By definition – as out of the total two outcomes Head and Tail, our favourable outcomes Head will occur only once.)
When coin A is tossed, the probability of getting two heads and one tail is
$P{(HHT)_A} = \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2}$
Therefore,
 $P{(HHT)_A} = \dfrac{1}{8}\;{\text{ }}.......{\text{(1)}}$
Coin B is the biased coin and
The probability of occurring head for coin A is $P{(H)_B} = \dfrac{2}{3}$ (Given)
Therefore, the probability of getting tail is equal to one minus the probability of getting Head
$
  P{(T)_B} = 1 - P(H){}_B \\
  P{(T)_B} = 1 - \dfrac{2}{3} \\
  P{(T)_B} = \dfrac{1}{3} \\
 $
Now, by substituting the values -
When coin B is tossed, the probability of getting two heads and one tail is
$P{(HHT)_B} = \dfrac{2}{3} \times \dfrac{2}{3} \times \dfrac{1}{3}$
Therefore, $P{(HHT)_B} = \dfrac{4}{{27}}\,{\text{ }}.......{\text{(2)}}$
Similarly, for coin C
The probability of occurring head for coin C is
$P{(H)_C} = \dfrac{1}{3}$ (Given)
Therefore, the probability of getting tail is equal to one minus the probability of getting Head
$
  P{(T)_B} = 1 - P(H){}_B \\
  P{(T)_B} = 1 - \dfrac{1}{3} \\
  P{(T)_B} = \dfrac{2}{3} \\
 $
Now, by substituting the values -
When coin C is tossed, the probability of getting two heads and one tail is
$P{(HHT)_C} = \dfrac{1}{3} \times \dfrac{1}{3} \times \dfrac{2}{3}$
Therefore, $P{(HHT)_C} = \dfrac{2}{{27}}\;{\text{ }}........{\text{(3)}}$
From the Equations $(1),\,{\text{(2) \& (3)}}$
Now, combined probability of all the three coins is –
$
  P{(HHT)_T} = P{(HHT)_A} + P{(HHT)_B} + P{(HHT)_C} \\
  P{(HHT)_T} = \dfrac{1}{8} + \dfrac{4}{{27}} + \dfrac{2}{{27}} \\
 $
Simplify by taking LCM
$
  P{(HHT)_T} = \dfrac{{27 + 32 + 16}}{{216}} \\
  P{(HHT)_T} = \dfrac{{75}}{{216}} \\
 $
(Take from both the numerator and the denominator; also same number from numerator and denominator cancels each other)
$P{(HHT)_T} = \dfrac{{25}}{{72}}$
Now, the probability that the chosen coin was A, giving two heads and one tail when tossed is –
$
  P(for{\text{ the coin A) = }}\dfrac{{P{{(HHT)}_A}}}{{P(HHT){}_T}} \\
  P(for{\text{ the coin A)}} = \dfrac{{\dfrac{1}{8}}}{{\dfrac{{25}}{{72}}}} \\
 $
Use the property, denominator or numerator goes in denominator and denominator’s denominator goes in numerator.
$\therefore P(for{\text{ the coin A) = }}\dfrac{{1 \times 72}}{{8 \times 25}}$
Now, simplify the right hand side of the equation –
$\therefore P(for{\text{ the coin A) = }}\dfrac{9}{{25}}$
Thus, the required answer is the probability that the chosen coin was A, giving two heads and one tail when tossed is –
$\therefore P(for{\text{ the coin A) = }}\dfrac{9}{{25}}$
Hence, from the given multiple choices, option A is the correct answer.

Note:
For this type of probability problems, just follow the general formula for probability and basic simplification properties for the fractions. Always remember the probability of any event lies between zero and one.