Question

Let U1 and U2 be two urns such that U1 contains 3 white and 2 red balls, and U2 contains only 1 white ball. A fair coin is tossed. If the head appears then 1 ball is drawn at random from U1 and put into U2. However, if the tail appears then 2 balls are drawn at random from U1 and put into U2. Now 1 ball is drawn at random from U2. Given that the drawn ball from U2 is white, the probability that head appeared on the coin is

1. $\dfrac{{17}}{{23}}$
2. $\dfrac{{11}}{{23}}$
3. $\dfrac{{15}}{{23}}$
4. $\dfrac{{12}}{{23}}$

Answer 1

Hint : Probability means possibility. Here, we are given U1 which contains 3 white and 2 red balls and U2 which has only one white ball. When a coin is tossed, according to that condition, we need to move the balls from U1 to U2 as given in the question. So first make the possible cases all both head and tail. And then, find the probability of each i.e. head and tail i.e. we will use the combination formula ${}^n{C_r} = \dfrac{{(n!)}}{{(r!)(n - r)!}}$. Lastly, find the probability when a ball drawn from U2 is white, when the head appears using the probability formula.

Complete step-by-step answer :
Given that, urn U1 contains 3 white and 2 red balls and urn U2 contains 1 white ball.
Total number of balls U1 has is 5.
So, the probability when one ball is drawn from U1 is $\dfrac{1}{5}$ .
Also, the total number of balls U2 has 1.
Now, the movement of balls from U1 to U2 on the condition that head or tail appears on the coin as shown below:
Let the events be defined as:
W: drawing a white ball from U2
H: tossing a head
T: tossing a tail
First, we toss a coin.
Here, P (H) = P (T) = $\dfrac{1}{2}$
Head appears:
Given that, one ball at random moved from U1 to U2.
Case 1: White ball is selected from U1 – Total 2 white balls in U2
Case 2: Red ball is selected from U1 – Total 2 balls in U2 i.e. 1 white and 1 red ball

Tail appears:
Given that, two balls at random are drawn from U1 to U2 (i.e. Total 3 balls in U2).
Case 1: 2 white balls are selected from U1 – 3 white balls in U2
Case 2: 2 red balls are selected from U1 – 1 white and 2 red balls in U2
Case 3: 1 white and 1 red ball is selected from U1 – 2 white and 1 red ball in U2
Case 4: 1 red and 1 white ball is selected from U1 – 2 white and 1 red ball in U2
Now we will solve this according to the given above explanation.
First,
When head is tossed:
The probability of drawing a white ball from U2 is
$P(\dfrac{W}{H}) = \dfrac{{{}^3{C_1}}}{{{}^5{C_1}}} \times \dfrac{{{}^2{C_1}}}{{{}^2{C_1}}} + \dfrac{{{}^2{C_1}}}{{{}^5{C_1}}} \times \dfrac{{{}^1{C_1}}}{{{}^2{C_1}}}$
$= \dfrac{3}{5} \times \dfrac{2}{2} + \dfrac{2}{5} \times \dfrac{1}{2}$
$= \dfrac{3}{5} \times 1 + \dfrac{1}{5}$
$= \dfrac{3}{5} + \dfrac{1}{5}$
$= \dfrac{4}{5}$
Next,
When tail is tossed:
The probability of drawing a white ball from U2 is
$P(\dfrac{W}{T}) = \dfrac{{{}^3{C_2}}}{{{}^5{C_2}}} \times \dfrac{{{}^3{C_1}}}{{{}^3{C_2}}} + \dfrac{{{}^2{C_2}}}{{{}^5{C_2}}} \times \dfrac{{{}^1{C_1}}}{{{}^3{C_1}}} + \dfrac{{{}^3{C_1} \times {}^2{C_1}}}{{{}^5{C_2}}} \times \dfrac{{{}^2{C_1}}}{{{}^3{C_2}}}$
$= \dfrac{{3 \times 2}}{{5 \times 4}} \times 1 + \dfrac{{1 \times 2}}{{5 \times 4}} \times \dfrac{1}{3} + \dfrac{{3 \times 2 \times 2}}{{5 \times 4}} \times \dfrac{2}{3}$
$= \dfrac{3}{{10}} + \dfrac{1}{{30}} + \dfrac{2}{5}$
$= \dfrac{{9 + 1 + 12}}{{30}}$
$= \dfrac{{22}}{{30}}$
Thus, when a ball drawn from U2 is white, then the probability that a head appears on the coin is
$P(\dfrac{H}{W}) = \dfrac{{P(H) \times P(\dfrac{W}{H})}}{{P(H) \times P(\dfrac{W}{H}) + P(T) \times P(\dfrac{W}{T})}}$
$= \dfrac{{\dfrac{1}{2} \times \dfrac{4}{5}}}{{\dfrac{1}{2} \times \dfrac{4}{5} + \dfrac{1}{2} \times \dfrac{{22}}{{30}}}}$
$= \dfrac{{\dfrac{2}{5}}}{{\dfrac{2}{5} + \dfrac{{11}}{{30}}}}$
$= \dfrac{{\dfrac{2}{5}}}{{\dfrac{{2(6) + 11}}{{30}}}}$
$= \dfrac{{\dfrac{2}{5}}}{{\dfrac{{12 + 11}}{{30}}}}$
$= \dfrac{{\dfrac{2}{5}}}{{\dfrac{{23}}{{30}}}}$
$= \dfrac{2}{5} \div \dfrac{{23}}{{30}}$
$= \dfrac{2}{5} \times \dfrac{{30}}{{23}}$
$= 2 \times \dfrac{6}{{23}}$
$= \dfrac{{12}}{{23}}$
So, the correct answer is “Option 4”.

Note : The probability formula is defined as the possibility of an event to happen is equal to the ratio of the number of favourable outcomes and the total number of outcomes. The probability of all the events in a sample space adds up to 1. Permutation and combination are the ways to represent a group of objects by selecting them in a set and forming subsets. When we select the data or objects from a certain group, it is said to be permutations, whereas the order in which they are represented is called combination.