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If a dice is rolled, what is the probability of getting a
(i) prime number
(ii) multiple of 3
(iii) number 7

seo-qna
Last updated date: 20th Jul 2024
Total views: 348.3k
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Answer
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Hint: When a dice is thrown, the possible outcomes are 1, 2, 3, 4, 5 and 6. The sample space will be $S=\left\{ 1,2,3,4,5,6 \right\}$ . Probability of an event E is given by the formula $P\left( E \right)=\dfrac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}$ , where total number of outcomes is the number of elements in the sample space. Prime numbers are the numbers that have factors as 1 and that number itself. To find the probability of getting a prime number, we have to check the number of prime numbers in the sample space and substitute the values in the formula for probability. A multiple is the product result of one number multiplied by another number. To find the probability of getting a multiple of 3, we have to check the number of multiples of 3 in the sample space and substitute the values in the formula for probability. A dice will never show a number 7 and above.

Complete step by step solution:
We know that a dice has six sides and each side has dots that represent numbers from 1 to 6. We know that when a dice is thrown, the possible outcomes are 1, 2, 3, 4, 5 and 6. We can write the sample space of this experiment as
$S=\left\{ 1,2,3,4,5,6 \right\}$
We know that probability of an event E is the number of favourable outcomes divided by the total number of outcomes.
$P\left( E \right)=\dfrac{\text{Number of favourable outcomes}}{\text{Total number of outcomes}}...\left( i \right)$
The total number of outcomes is the number of elements in the sample space. Here, we can see that the total number of outcomes = 6.
Now, let us find the probability of each case.
(i) We have to find the probability of getting a prime number. We know that prime numbers are the numbers that have factors as 1 and that number itself. From the sample space, S we can see that 2, 3 and 5 are the prime numbers.
Therefore, number of favourable outcomes = 3
We can find the probability of getting a prime number by substituting these values in the formula (i).
$\Rightarrow \text{P}\left( \text{prime number} \right)=\dfrac{3}{6}$
We have to simplify the result by cancelling the common factor 3.
$\Rightarrow \text{P}\left( \text{prime number} \right)=\dfrac{1}{2}$
Hence, the probability of getting a prime number is $\dfrac{1}{2}$ .
(ii) We have to find the probability of getting a multiple of 3. We know that a multiple is the product result of one number multiplied by another number. We also know that multiples of 3 are 3, 6, 9, 12, … . From the sample space, we can see that the multiples of 3 are 3 and 6.
Therefore, the number of favourable outcomes = 2.
We can find the probability of getting a multiple of 3 by substituting these values in the formula (i).
$\Rightarrow \text{P}\left( \text{multiple of 3} \right)=\dfrac{2}{6}$
We have to simplify the above result by cancelling the common factor 2.
$\Rightarrow \text{P}\left( \text{multiple of 3} \right)=\dfrac{1}{3}$
Hence, the probability of getting a multiple of 3 is $\dfrac{1}{3}$ .
(iii) We have to find the probability of getting a number 7. From the sample space, we can see that a number 7 and above will never appear on a dice.
Therefore the number of favourable outcomes = 0.
We can find the probability of getting a number 7 by substituting these values in the formula (i).
$\begin{align}
  & \Rightarrow \text{P}\left( \text{number 7} \right)=\dfrac{0}{6} \\
 & \Rightarrow \text{P}\left( \text{number 7} \right)=0 \\
\end{align}$
Hence, the probability of getting a number 7 is 0.

Note: Students have a chance of making mistake by writing the formula for probability as $P\left( E \right)=\dfrac{\text{Total number of outcomes}}{\text{Number of favourable outcomes}}$ . They must read the question clearly and create a sample space accordingly. Any mistake in the sample space will make the entire solution wrong. Students must note that the probability of an event cannot be negative and greater than 1, that is, it falls in the range $0\le P\left( E \right)\le 1$ .