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In a class of 60 students 30 opted for Mathematics, 32 opted for Biology and 24 opted for both Mathematics and Biology. If one of these students is selected at random, find the probability that:
i) The students opted for Mathematics or Biology.
ii) The students opted for neither Mathematics nor Biology.
iii) The students opted for Mathematics but not Biology.

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Last updated date: 16th Jun 2024
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Answer
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Hint: We have to find the probability for the required attributes. We will solve the problem by substituting the given values into some relation of probability.

Formula used:
The probability formula is defined as the possibility of an event to happen is equal to the ratio of the number of favorable outcomes and the total number of outcomes.
We know that,
\[n(M \cup B) = n(M) + n(B) + n(M \cap B)\]
Here, number of students who opted for Mathematics is \[n(M)\].
The number of students who opted for Biology is \[n(B)\].
The number of students who opted for both Mathematics and Biology is \[n(M \cap B)\].

Complete step by step answer:
It is given that the total number of students in the class is \[60\].
The number of students who opted for Mathematics is \[30\].
The number of students who opted for Biology is \[32\].
The number of students who opted for both Mathematics and Biology is \[24\].
We have to find the probability that:
(i) The students opted for Mathematics or Biology.
(ii) The students opted for neither Mathematics nor Biology.
(iii) The students opted for Mathematics but not Biology.
So, as per the given information
 \[n(M) = 30\]
\[n(B) = 32\]
\[n(M \cap B) = 24\]
The number of students who opted for Mathematics or Biology is \[n(M \cup B)\].
We know that,
\[n(M \cup B) = n(M) + n(B) + n(M \cap B)\]
Substitute the values in the above formula we get,
\[\Rightarrow n(M \cup B) = 30 + 32 - 24\]
Simplifying we get,
\[\Rightarrow n(M \cup B) = 38\]
(i) The students opted for Mathematics or Biology.
So, the probability of students opted for Mathematics or Biology is \[\dfrac{{38}}{{60}}\]
Simplifying we get,
$\therefore $ The probability of students opted for Mathematics or Biology is \[ = \dfrac{{19}}{{30}}\].

(ii) The students opted for neither Mathematics nor Biology.
We know that the total probability of any event is 1.
So, the probability of students opted for neither Mathematics nor Biology is \[ = 1 - \dfrac{{19}}{{30}}\]
Simplifying we get,
$\therefore $ The probability of students opted for neither Mathematics nor Biology is \[ = \dfrac{{11}}{{30}}\].

(iii) The students opted for Mathematics but not Biology.
The number of students opted for only Mathematics but not Biology is \[ = n(M) - n(M \cap B)\]
Substitute the values we get,
The number of students opted for only Mathematics but not Biology is \[ = 30 - 24 = 6\]
$\therefore $ The probability of students opted for only Mathematics but not Biology is \[ = \dfrac{6}{{60}} = \dfrac{1}{{10}}\].

Note:
In this, we are finding the probability of the required thing. For that, we have used some probability relations. By substituting the given values into that probability relations, we got the required correct results. So students may go wrong on that substitutions. They have to focus on that calculation.