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from the given set $\left\{ 2,4,6,8,10,12,14,16,18 \right\}$ , if a number is selected at
random, calculate the probability of the number being divisible by 3.
(a) $\dfrac{1}{3}$
(b) $\dfrac{4}{9}$
(c) $\dfrac{5}{9}$
(d) $\dfrac{2}{3}$
(e) 1

seo-qna
Last updated date: 28th Mar 2024
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MVSAT 2024
Answer
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Hint: First find out the numbers which are divisible by 3.
Then divide the number of favourable outcomes by the total number of outcomes to get the required probability.

Complete step by step solution:
Probability means possibility. Probability is a measure of the likelihood of an event to occur. The probability formula is defined as the possibility of an event to happen is equal to the ratio of the number of favourable outcomes and the total number of outcomes.
Here in the given question we have a set $\left\{ 2,4,6,8,10,12,14,16,18 \right\}$
The set has 9 elements.
So if we select a number from this set at random we can get at most 9 different numbers.
So, here the total number of outcomes are 9.
Now we have to find out the total number of favourable outcomes.
Here, our favourable outcomes will be those numbers which are divisible by 3.
Divisible by 3 means, if we divide that number by 3 the remainder should be 0.
Now let us check which numbers are divisible by 3 here.
We know that, 2 is not divisible by 3.
4 is not divisible by 3, as the remainder is 1.
6 is divisible by 3, as the remainder is 0.
8 is not divisible by 3, as the remainder is 2.
10 is not divisible by 3, as the remainder is 1.
12 is divisible by 3, as the remainder is 0.
14 is not divisible by 3, as the remainder is 2.
16 is not divisible by 3, as the remainder is 1.
18 is divisible by 3, as the remainder is 0.

Therefore, there are 3 numbers which are divisible by 3.
Those are 6, 12, 18.
So, our total number of favourable outcomes are 3.
So, the probability of selecting the numbers which are divisible by 3 is:
$\dfrac{3}{9}=\dfrac{1}{3}$
Hence, option (a) is correct.

Note: We generally make mistakes while counting the total number of favourable cases. We have to read the question very carefully to understand the favourable cases.
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