Question

A box contains 90 discs numbered from 1 to 90. If one disc is drawn from the box at random, find the probability that it bears a prime number less than 23.
A) $$\dfrac{7}{{90}}$$
B) $$\dfrac{{10}}{{90}}$$
C) $$\dfrac{4}{{45}}$$
D) $$\dfrac{9}{{89}}$$

Answer 1

Hint: We will first find all the prime numbers between 1 and 23 and calculate the frequency of such numbers. After that we will divide this number by 90 to get the required probability.

Complete step-by-step answer:
Let us first of all understand what do we mean by probability:
Probability is simply how likely something is to happen. Whenever we're unsure about the outcome of an event, we can talk about the probabilities of certain outcomes—how likely they are.
If we need to find the probability of a specific event E, when we have S as the total possible outcomes, then probability of E is given by $P(E) = \dfrac{{n(E)}}{{n(S)}}$, where P(E) is the required probability, n(E) is the number or outcomes in E and n(S) is the total number of possible outcomes in the experiment assuming each event is equally likely to happen.
Now, to solve the given question, we will have to find prime numbers less than 23 and greater than 1.
A prime number is an integer which only has two possible factors which are 1 and the number itself.
Hence, the prime numbers from 1 to 23 are 2, 3, 5, 7, 11, 13, 17, 19.
We see that we have 8 prime numbers between 1 and 23.
So, if we say E = event that when we get out a random disc, it bears a prime number less than 23, then n(E) = 8.
Now, we have 90 numbers to pull out of the box. Hence, the total number of outcomes are 90.
Hence, n(S) = 90
Now applying the formula: $P(E) = \dfrac{{n(E)}}{{n(S)}}$, where P(E) is the required probability, n(E) is the number or outcomes in E and n(S) is the total number of possible outcomes in the experiment assuming each event is equally likely to happen.
We will have:- $P(E) = \dfrac{{n(E)}}{{n(S)}} = \dfrac{8}{{90}}$.
We see that there is no such option.
So simplifying it further, we have:
$P(E) = \dfrac{8}{{90}} = \dfrac{4}{{45}}$
Hence, we have $P(E) = \dfrac{4}{{45}}$.

Hence, the correct option is (C).

Note: The students might make the mistake of always simplifying the answer before matching the non simpler form to one of the options, which might be there in the options and thus not getting the answer as required. So, you must take care of matching that to options as well before simplifying it.
The students might include 1 or 23 or both in prime numbers but always keep in mind that 1 is neither prime nor composite, so we cannot have 1 and we need numbers strictly less than 23, so 23 cannot be there as well.