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A die is numbered in such a way that its faces show the number 1, 2, 2, 3, 3, 6. It is thrown two times and the total score in two throws is noted. Complete the following table which gives a few values of the total score on the two throws:
What is the probability that the total score is
(i) even?
(ii) 6?
(iii) at least 6?"
Given:
A die is numbered in such a way that its faces show the number 1, 2, 2, 3, 3, 6. It is thrown two times and the total score in two throws is noted.
To do:
We have to find the probability that the total score is
(i) even
(ii) 6
(iii) at least 6
Solution:
To complete the given table we just have to add the numbers on the faces.
When two dice are thrown, the total possible outcomes are $6\times6=36$.
All the possible outcomes are $(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4),$
$(2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1),$
$(5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)$
This implies,
The total number of possible outcomes $n=36$
(i) The number of outcomes where the total score is even $=18$
Total number of favourable outcomes $=18$
Probability of an event $=\frac{Number\ of\ favourable\ outcomes}{Total\ number\ of\ possible\ outcomes}$
Therefore,
Probability that the total score is even $=\frac{18}{36}$
$=\frac{1}{2}$
The probability of obtaining a total score that is even is $\frac{1}{2}$.
(ii) The number of outcomes where the total score is 6 $=4$
Total number of favourable outcomes $=4$
Probability of an event $=\frac{Number\ of\ favourable\ outcomes}{Total\ number\ of\ possible\ outcomes}$
Therefore,
Probability that the total score is 6 $=\frac{4}{36}$
$=\frac{1}{9}$
The probability of obtaining a total score that is 6 is $\frac{1}{9}$.
(iii) The number of outcomes where the total score is at least 6 $=15$
Total number of favourable outcomes $=15$
Probability of an event $=\frac{Number\ of\ favourable\ outcomes}{Total\ number\ of\ possible\ outcomes}$
Therefore,
Probability that the total score is at least 6 $=\frac{15}{36}$
$=\frac{5}{12}$
The probability of obtaining a total score that is at least 6 is $\frac{5}{12}$.
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