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An urn contains 5 red and 5 black balls. A ball is drawn at random; its color is noted and is returned to the urn. Moreover, 2 additional balls of the color drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red?

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Last updated date: 20th Jun 2024
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Answer
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Hint: First find the probability of the first ball drawn is red. Then add 2 red balls in an urn and again find the probability of the second ball drawn is red. After that multiply both to get the probability that both balls are drawn are red. Now, find the probability of the first ball drawn is black. Then add 2 black balls in an urn and again find the probability of the second ball drawn is red. After that multiply both to get the probability that the first ball is drawn is black and the second ball drawn is red. Now add both the probability to get the probability that the second ball is red.

Complete step by step answer:
Given: 5 red balls and 5 black balls
Let a red ball be drawn in the first attempt. Then,
$P\left( {{R}_{1}} \right)=\dfrac{5}{10}$
Cancel out the common factors,
$P\left( {{R}_{1}} \right)=\dfrac{1}{2}$
Now, two red balls are added to the urn, then the urn contains 7 red and 5 black balls.
$P\left( {{R}_{2}} \right)=\dfrac{7}{12}$
Now, the probability of drawing red balls on each attempt is,
$\Rightarrow P\left( {{R}_{1}}{{R}_{2}} \right)=P\left( {{R}_{1}} \right)\times P\left( {{R}_{2}} \right)$
Substitute the values,
$\Rightarrow P\left( {{R}_{1}}{{R}_{2}} \right)=\dfrac{1}{2}\times \dfrac{7}{12}$
Multiply the terms,
$\Rightarrow P\left( {{R}_{1}}{{R}_{2}} \right)=\dfrac{7}{24}$ …………..….. (1)
Let a black ball be drawn in the first attempt. Then,
$P\left( {{B}_{1}} \right)=\dfrac{5}{10}$
Cancel out the common factors,
$\Rightarrow P\left( {{B}_{1}} \right)=\dfrac{1}{2}$
Now, two black balls are added to the urn, then the urn contains 5 red and 7 black balls.
$\Rightarrow P\left( {{R}_{2}} \right)=\dfrac{5}{12}$
Now, the probability of drawing black balls on the first attempt and red ball on the second attempt is,
$\Rightarrow P\left( {{B}_{1}}{{R}_{2}} \right)=P\left( {{B}_{1}} \right)\times P\left( {{R}_{2}} \right)$
Substitute the values,
$\Rightarrow P\left( {{B}_{1}}{{R}_{2}} \right)=\dfrac{1}{2}\times \dfrac{5}{12}$
Multiply the terms,
$\Rightarrow P\left( {{B}_{1}}{{R}_{2}} \right)=\dfrac{5}{24}$ …………………...….. (2)
Therefore, the probability of drawing the second ball as red is
$\Rightarrow P\left( {{R}_{2}} \right)=P\left( {{R}_{1}}{{R}_{2}} \right)\times P\left( {{B}_{1}}{{R}_{2}} \right)$
Substitute the values from the equation (1) and (2)
$\Rightarrow P\left( {{R}_{2}} \right)=\dfrac{7}{24}+\dfrac{5}{24}$
Since the denomination of both the rational number is the same. So, add the numerator,
$\Rightarrow P\left( {{R}_{2}} \right)=\dfrac{12}{24}$
Cancel out the common factors,
$\Rightarrow P\left( {{R}_{2}} \right)=\dfrac{1}{2}$

Hence, the probability that the second ball is red is $\dfrac{1}{2}$.

Note:
Probability is a measure of the likelihood of an event to occur. Many events cannot be predicted with total certainty. We can predict only the chance of an event to occur i.e. how likely they are to happen, using it.
The probability formula is defined as the possibility of an event to happen is equal to the ratio of the number of favorable outcomes and the total number of outcomes.
$P\left( E \right)=\dfrac{n\left( A \right)}{n\left( S \right)}$
A probability of 0 means that an event is impossible.
A probability of 1 means that an event is certain.
An event with a higher probability is more likely to occur.
Probabilities are always between 0 and 1.