In a single throw of a die, the probability of getting a multiple of 3...
Given : A die is thrown once .
A die has 6 faces marked as 1, 2, 3, 4, 5 and 6.
If we throw one die then there possible outcomes are as follows: 1, 2, 3, 4, 5 and 6
Number of possible outcomes are = 6
Let E = Event of getting a getting a multiple of 3
Multiples of 3 are = 3, 6
Number of outcome favourable to E = 2
Probability (E) = Number of favourable outcomes / Total number of outcomes
P(E) = 2/6 = 1/3
Hence, the probability of getting a multiple of 3, P(E) = 1/3
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In a single throw of a die, the probability of getting a multiple of 3...
Just make sets of 3 numbers 1,2,3,4,5,6,7,8,9,... Means 1,2,3 4,5,6 7,8,9 In every set you will find only single number which is divisible by 3 In 1,2,3 3 is the number In 4,5,6 6 is the number In 7,8,9 9 is the number Likewise therefore, set contains total of 3 numbers but event is only 1. P(3) = 1/3.
In a single throw of a die, the probability of getting a multiple of 3...
B