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# A dice is thrown twice and the sum of the numbers appearing is observed to be 7. Find the conditional probability that the number 3 has appeared at least once. Verified
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Hint: Count the number of cases such that the sum of the number is 7 and find the probability of occurrence of 3 out of those cases.

Given, a dice is thrown twice and the sum of the numbers is 7. We need to find the conditional probability of occurrence of 3 at least once. The six-faced dice has one number on each face. They are 1,2,3,4,5 and 6. We need to find the combination of two numbers such that the sum is 7 to find the total number of cases. So, on observing, start with 1, 1+6 will give 7.

Similarly, 6 + 1 will also give the sum as 7. Next number is 2, 2+5 and 5+2 is 7. Similarly, 3 + 4 and 4+3. After this, we will get the repeated combination. So, we have (1,6), (6,1), (2,5), (5,2), (3,4) and (4,3). Total number of cases for the event to occur is 6 i.e. there are a total 6 ways so that sum would be 7. Now, the number of cases in which the number 3 appears at least once are (3,4) and (4,3) i.e. 2. The question is asking to find the conditional probability of the event (3 appearing at least once). The conditional probability is the probability of occurring one event with some relationship to one or more other events. So, occurrence of 3 at least once is one event, and it is occurring when the other event i.e. the sum to be 7 has already occurred.

So, the conditional probability = $\dfrac{{{\text{no}}{\text{. of cases when the numbered 3 occured at least once}}}}{{{\text{total number of cases when the sum is 7}}}}$
=$\dfrac{2}{6}$
=$\dfrac{1}{3}$
Hence, the required answer is $\dfrac{1}{3}$.

Note:- In these types of questions, the key concept is to have a good understanding of the conditional probability. It is the probability of occurrence of one event when another event has already occurred.
Last updated date: 21st Sep 2023
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