Question

A bag contains 5 balls of unknown colors. A ball is drawn at random from it and is found to be white. The probability that bag contains only white ball is:
A. \[\dfrac{3}{5} \]
B. \[\dfrac{1}{5} \]
C. \[\dfrac{2}{3} \]
D. \[\dfrac{1}{3} \]

Answer 1

Hint: Total number of balls is 5. Take events A, B, C, D, E as the bag contains 1, 2, 3, 4, 5 white balls respectively. Take Q as the event that balls drawn are white. Thus find \[P\left( {}^{E}/{}_{Q} \right)\] using Bayes theorem.

Complete step-by-step answer:
It is said that a bag contains 5 balls of unknown colors. So the total number of balls in the bag = 5.
Let’s take Q as the event that the drawn balls are white.
We need to find the probability that the bag contains only white balls, which means that all the 5 balls can be white.
Let A be the event that the bag contains 1 white ball.
Let B be the event that the bag contains 2 white balls. Similarly, let C be the event that the bag contains 3 white balls.
Let D be the event that the bag contains 4 white balls and let E be the event that the bag contains 5 white balls.
Thus we can say that the \[P\left( A \right)={}^{1}/{}_{5}\], i.e. the probability of occurrence of event A. Similarly, probability of occurrence of event B, \[P\left( B \right)={}^{1}/{}_{5}\]. Similarly, we can write that \[P\left( C \right)={}^{1}/{}_{5},\] \[P\left( D \right)={}^{1}/{}_{5},\] \[P\left( E \right)={}^{1}/{}_{5}\].
Now let us find \[P\left( {}^{Q}/{}_{A} \right)\] = probability of the event that the balls drawn are white and in event A, the bag contains 1 white ball.
\[P\left( {}^{Q}/{}_{A} \right)=\dfrac{^{1}{{C}_{1}}}{^{5}{{C}_{1}}}=\dfrac{1}{5}\].
Hence it is of the form, \[^{n}{{C}_{r}}{{=}^{5}}{{C}_{1}}=\dfrac{5!}{(5-1)!1!}=\dfrac{5!}{4!1!}=\dfrac{5\times 4!}{4!\times 1!}=5\].
Similarly, \[P\left( {}^{Q}/{}_{A} \right)\] = the event that the bag contains 2 white balls.
\[P\left( {}^{Q}/{}_{B} \right)=\dfrac{^{2}{{C}_{1}}}{^{5}{{C}_{1}}}=\dfrac{2}{5}\].
\[\begin{align}
  & P\left( {}^{Q}/{}_{C} \right)=\dfrac{^{3}{{C}_{1}}}{^{5}{{C}_{1}}}=\dfrac{3}{5} \\
 & P\left( {}^{Q}/{}_{D} \right)=\dfrac{^{4}{{C}_{1}}}{^{5}{{C}_{1}}}=\dfrac{4}{5} \\
 & P\left( {}^{Q}/{}_{E} \right)=\dfrac{^{5}{{C}_{1}}}{^{5}{{C}_{1}}}=1 \\
\end{align}\]
Thus, we got,
\[P\left( A \right)=P\left( B \right)=P\left( C \right)=P\left( D \right)=P\left( E \right)=\dfrac{3}{5} \]

\[P\left( {}^{Q}/{}_{A} \right)={}^{1}/{}_{5},P\left( {}^{Q}/{}_{B} \right)={}^{2}/{}_{5},P\left( {}^{Q}/{}_{C} \right)={}^{3}/{}_{5},P\left( {}^{Q}/{}_{D} \right)={}^{4}/{}_{5},P\left( {}^{Q}/{}_{E} \right)=1\]
Now let us use Bayes theorem to find the value of \[P\left( {}^{E}/{}_{Q} \right)\].
\[P\left( {}^{E}/{}_{Q} \right)=\dfrac{{}^{1}/{}_{5} \times {}^{5}/{}_{5}}{\left( \dfrac{1}{5}\times \dfrac{1}{5} \right)+\left( \dfrac{1}{5}\times \dfrac{2}{5} \right)+\left( \dfrac{1}{5}\times \dfrac{3}{5} \right)+\left( \dfrac{1}{5}\times \dfrac{4}{5} \right)+\left( \dfrac{1}{5}\times \dfrac{5}{5} \right)}=\dfrac{{}^{5}/{}_{25}}{\dfrac{1}{25}\left( 1+2+3+4+5 \right)}\]
Cancel out \[{}^{1}/{}_{25}\] from the numerator and the denominator.
We get,
\[P\left( {}^{E}/{}_{Q} \right)=\dfrac{5}{1+2+3+4+5}=\dfrac{5}{15}=\dfrac{1}{3}\].
Thus we got the probability that the bag contains only white balls as \[\dfrac{1}{3}\].
Option D is the correct answer.

Note: Bayes theorem provides a way to revise existing predictions or theories given new or additional evidence. The Bayes theorem describes the probability of an event based on prior knowledge of conditions.