Question

# A bag contains tickets numbered from 1 to 20. Two tickets are drawn. Find the probability that on one there is a prime number and on the other, there is a multiple of 4.

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Hint: To solve this problem we need to find the favorable cases for the given numbers.Here we have to find the favourable cases for both prime numbers and multiples of 4.
Required formula
Probability of event to happen $P(E)$
$= \dfrac{{Number\,of\,favourable\,outcome}}{{total\,number\,of\,outcomes}}$

Complete step-by- step solution:
Given that,
A bag contain ticket numbered from 1 to 20
The two tickets are drawn
1 2 3 4 5 6 7 8 9 10
11 12 13 14 15 16 17 18 19 20
Now, prime numbers are the numbers that have only two factors i.e. 1 and itself.
Here prime numbers are - 1 2 3 5 7 11 13 17 i.e. 8 favorable cases.
Here multiples of 4 are$\to$ 4, 8, 12, 16, 20 i.e. 5 favorable cases
Total numbers $= 20$
Now probability of prime number $= \dfrac{{Number\,of\,favourable\,outcome}}{{total\,number\,of\,outcomes}}$$= \dfrac{8}{{20}}$
Probability of multiple of 4 $=\dfrac{{Number\,of\,favourable\,outcome}}{{total\,number\,of\,outcomes}}$$= \dfrac{5}{{20}}$
We need to find Probability (one there is a prime number and other there is a multiple of$4$)
$= \dfrac{8}{{20}} \times \dfrac{5}{{19}} + \dfrac{5}{{20}} \times \dfrac{8}{{19}}$
$= \dfrac{{40 + 40}}{{20 \times 19}} = \dfrac{{80}}{{20 \times 19}} = \dfrac{4}{{19}}$
Probability (one there is prime number and another is multiple of$4$)
$= \dfrac{4}{{19}}$

Note: Probability means possibility.
We must remember that when we have picked one number out of 20 then while picking the second number from the same, the total possibilities will be 19.
We need to remember that in probability we use multiplication until the case is completed but we use addition when the case has been completed and it is ‘either or case’.
We use the basic concept of probability.