Question

Given 3 identical boxes I, II and III each containing two coins. In box I both are gold coins; in box II both are silver coins and in box III one is gold and one is silver. A coin is drawn from one of the boxes. If the coin is of gold, what is the probability that the other coin is the box is also of gold. (write up to 2 decimal places).

Answer 1

Hint: First assume a variable for each event like a box having 2 gold, 2 silver, I gold one silver and 2nd coin gold. Now find the probability of each of the first three events where the 4th event is given. For these, apply the conditional probability concept. First find the probability of each of the first three events. Next, find the probability of the 4th event given each of the first three events. By following formula find required result: for A, B, C, D 4 events.
$P\left( A/D \right)=\dfrac{P\left( A \right)\times P\left( D/A \right)}{P\left( A \right)\times P\left( D/A \right)+P\left( B \right)\times P\left( D/B \right)+P\left( C \right)\times P\left( D/C \right)}$

Complete step-by-step answer:
Let us assume the first event variable A, denoting that: A- event of selecting box 1 having two gold coins in it. Let us assume the second event variable B, denoting that: B-event of selecting box 2 having 2 silver coins in it. Let us assume the third event variable C, denoting that: C-event of selecting box 3 having 1gold 1 silver coins in it. Let us assume the fourth event variable D, denoting that: D-event of the second coin in the bag being a gold coin.
We need the probability of another coin being gold if one the first coin is gold. In other words, we need to select box 1.
i.e. $P\left( A/D \right)$
By basic probability knowledge, we can say the formula:
$P\left( A/D \right)=\dfrac{P\left( A \right)\times P\left( D/A \right)}{P\left( A \right)\times P\left( D/A \right)+P\left( B \right)\times P\left( D/B \right)+P\left( C \right)\times P\left( D/C \right)}$
By the given condition, the boxes are equally likely to be selected. So, probabilities of each box are equal. So, dividing 1 to 3 equal parts we can say that the probabilities:
$P\left( A \right)=P\left( B \right)=P\left( C \right)=\dfrac{1}{3}$
Probability that second coin is gold coin in the box 1 is 1 because both are gold:
$P\left( G/A \right)=1$
Probability that second coin is gold coin in box 2 is 0 as both are silver:
$P\left( G/B \right)=0$
Probability that the second is gold in box 3 is $\dfrac{1}{2}$ as it may be gold or silver.
$P\left( G/C \right)=\dfrac{1}{2}$
By substituting all these values back into the formula, we get:
$P\left( A/D \right)=\dfrac{1\times \dfrac{1}{3}}{\dfrac{1}{3}\times 1+\dfrac{1}{3}\times 0+\dfrac{1}{3}\times \dfrac{1}{2}}$
By cancelling the common term in the above fraction, we get:
$P\left( A/D \right)=\dfrac{1}{1+\dfrac{1}{2}}$
By simplifying the above equation, we get the value as:
$P\left( A/D \right)=\dfrac{2}{3}$
Therefore, the probability that if we draw one gold coin second draw will also be gold coin is $\dfrac{2}{3}=0.66$

Note: The above observation that the required result means that selecting box 1 is very important. While substituting in the formula we can miss any term. So, do the substitution step carefully. Whenever you see the term identical in question that means that they are describing equally likeliness of a particular event. We wrote 0.66 instead of fraction because they asked in question also $\dfrac{3}{2}$ is correct.