
A bag contains 18 balls out of which x balls are red.
If 2 more red balls are put in the bag then the probability of drawing a red ball will be $\dfrac{9}{8}$ times the probability of drawing in the first case. Find the value of x.
Answer
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Hint: Probability is a measure of the likelihood that an event will happen.
\[Probability{\text{ }} = \;\dfrac{{favorable\;outcomes}}{{possible\;outcomes}}\]
We will use the concept of probability to solve this problem.
Complete step-by-step answer:
The total number of elementary events of a trial are known as the total number of cases.
Bag contains =18 balls
Desired outcome of an elementary event is called a favourable event.
Number of red balls =x
Probability of red ball coming in first case = \[\dfrac{{Favourable\;ball}}{{total\;no\;of\;ball}} = \dfrac{x}{{18}}..................(1)\]
If two more red ball add
Since, the selected ball has to be red then this is our favourable event. Number of red ball \[ = x + 2\]
Total number of balls \[ = 18 + 2 = 20\]
Probability of red ball in second case = \[\dfrac{{Favourable\;ball}}{{total\;no\;of\;ball}} = \dfrac{{x + 2}}{{20}}..................(2)\]
As per given condition
Probability of red ball in second case = $\dfrac{9}{8}$ Probability of red ball coming in first case
By using equation (1) and (2) in above equation
\[
\Rightarrow \dfrac{{x + 2}}{{20}} = \dfrac{9}{8}\left( {\dfrac{x}{{18}}} \right) \\
\Rightarrow \dfrac{{x + 2}}{5} = \dfrac{1}{2}\left( {\dfrac{x}{2}} \right) \\
\Rightarrow 4\left( {x + 2} \right) = 5x \\
\Rightarrow 4x + 8 = 5x \\
\Rightarrow 5x - 4x = 8 \\
\Rightarrow x = 8 \\
\]
Therefore, the initial number of red balls in the bag is equal to 8.
Note: If all outcomes are favorable for a certain event, its probability is 1. For example, the probability of rolling a 6 or lower on one die is = 1.
If none of the possible outcomes are favorable for a certain event (a favorable outcome is impossible), the probability is 0. For example, the probability of rolling a 7 on one die is = 0.
\[Probability{\text{ }} = \;\dfrac{{favorable\;outcomes}}{{possible\;outcomes}}\]
We will use the concept of probability to solve this problem.
Complete step-by-step answer:
The total number of elementary events of a trial are known as the total number of cases.
Bag contains =18 balls
Desired outcome of an elementary event is called a favourable event.
Number of red balls =x
Probability of red ball coming in first case = \[\dfrac{{Favourable\;ball}}{{total\;no\;of\;ball}} = \dfrac{x}{{18}}..................(1)\]
If two more red ball add
Since, the selected ball has to be red then this is our favourable event. Number of red ball \[ = x + 2\]
Total number of balls \[ = 18 + 2 = 20\]
Probability of red ball in second case = \[\dfrac{{Favourable\;ball}}{{total\;no\;of\;ball}} = \dfrac{{x + 2}}{{20}}..................(2)\]
As per given condition
Probability of red ball in second case = $\dfrac{9}{8}$ Probability of red ball coming in first case
By using equation (1) and (2) in above equation
\[
\Rightarrow \dfrac{{x + 2}}{{20}} = \dfrac{9}{8}\left( {\dfrac{x}{{18}}} \right) \\
\Rightarrow \dfrac{{x + 2}}{5} = \dfrac{1}{2}\left( {\dfrac{x}{2}} \right) \\
\Rightarrow 4\left( {x + 2} \right) = 5x \\
\Rightarrow 4x + 8 = 5x \\
\Rightarrow 5x - 4x = 8 \\
\Rightarrow x = 8 \\
\]
Therefore, the initial number of red balls in the bag is equal to 8.
Note: If all outcomes are favorable for a certain event, its probability is 1. For example, the probability of rolling a 6 or lower on one die is = 1.
If none of the possible outcomes are favorable for a certain event (a favorable outcome is impossible), the probability is 0. For example, the probability of rolling a 7 on one die is = 0.
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