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# A bag contains 30 balls numbered from 1 to 30. One ball is drawn at random.A. The probability that the number of the drawn ball will be a multiple of 3 is $\dfrac{1}{3}$ B. The probability that the number of the drawn ball will be a multiple of 2 is 0.5C. The probability that the number of the drawn ball will be a multiple of 7 is $\dfrac{2}{15}$ D. The probability that the number of the drawn ball will be a multiple of 3 and 5 is $\dfrac{2}{15}$

Last updated date: 04th Aug 2024
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Hint: The probability of an event is equal to the number of favourable outcomes divided by the total number of outcomes of that event. It is basically a ratio of the two calculated in order to know about the possibility of occurrence of an event. Thus, we shall first write all the sample space of drawing a ball at random and then we will compute the probability of the four given events.

Complete step by step solution:
The event is that one ball is drawn at random from 30 balls numbered from 1 to 30.
The sample space of the event, $S$ is $\left\{ 1,2,3,4............,28,29,30 \right\}$. Thus, the total number of outcomes of this event is 30.
Now, we shall find the favourable outcomes and find the probability for each option in the problem.
The favourable outcomes for the number on the ball being multiple of 3, ${{E}_{3}}$ are $\left\{ 3,6,9,12,15,18,21,24,27,30 \right\}$ which is equal to 10.
Thus, the probability of the number being multiple of 3, $P\left( 3 \right)$ is given as:
$P\left( 3 \right)=\dfrac{{{E}_{3}}}{S}$
$\Rightarrow P\left( 3 \right)=\dfrac{10}{30}$
$\Rightarrow P\left( 3 \right)=\dfrac{1}{3}$ ……………….. (1)
The favourable outcomes for the number on the ball being multiple of 5, ${{E}_{5}}$ are $\left\{ 5,10,15,20,25,30 \right\}$ which is equal to 6.
Thus, the probability of the number being multiple of 5, $P\left( 5 \right)$ is given as:
$P\left( 5 \right)=\dfrac{{{E}_{5}}}{S}$
$\Rightarrow P\left( 5 \right)=\dfrac{6}{30}$
$\Rightarrow P\left( 5 \right)=\dfrac{1}{5}$
$\Rightarrow P\left( 5 \right)=0.2$ …………………… (2)
The favourable outcomes for the number on the ball being multiple of 7, ${{E}_{7}}$ are $\left\{ 7,14,21,28 \right\}$ which is equal to 4.
Thus, the probability of the number being multiple of 7, $P\left( 7 \right)$ is given as:
$P\left( 7 \right)=\dfrac{{{E}_{7}}}{S}$
$\Rightarrow P\left( 7 \right)=\dfrac{4}{30}$
$\Rightarrow P\left( 7 \right)=\dfrac{2}{15}$ …………………… (3)
The favourable outcomes for the number on the ball being multiple of 3 and 5, ${{E}_{3,5}}$ are $\left\{ 15,30 \right\}$ which is equal to 2.
Thus, the probability of the number being multiple of 3 and 5, $P\left( 3,5 \right)$ is given as:
$P\left( 3,5 \right)=\dfrac{{{E}_{3,5}}}{S}$
$\Rightarrow P\left( 3,5 \right)=\dfrac{2}{30}$
$\Rightarrow P\left( 3,5 \right)=\dfrac{1}{15}$ …………………… (4)

So, the correct answer is “Option A, B and C”.

Note: The sample space of an event is the total number of outcomes of an event. Before calculating the probability of an event, we must carefully analyze and write all the possible outcomes. In more complex like dice probability problems, the sample space becomes trickier.