A pair of the die is thrown. Find the probability of getting a sum of 10 or more, if 5 appears on the first die.
In throwing a dice, total possible outcomes
= 1,2,3,4,5,6
For two dice, n(S) = 6×6 = 36
So, n(S)=6
Favorable cases where the sum is 10 or more with 5 on 1st die =(5,5),(5,6)
Event of getting the sum is 10 or more with 5 on 1st die = n(E) =2
Hence, the probability of getting a sum of 10 or more with 5 on 1st die
[tex] = \frac{n(E)}{ n(S)} [/tex]
[tex] = \frac{2}{36} [/tex]
[tex] = \frac{1}{18} [/tex]
[tex]\huge\bf\underbrace{\pink{A}\orange{n}\blue{s}\red{w}\green{e}\purple{r}}[/tex]
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Question-;
A pair of the die is thrown. Find the probability of getting a sum of 10 or more, if 5 appears on the first die.
SOLUTION-;
In throwing a dice, total possible outcomes
= 1,2,3,4,5,6
For two dice, n(S) = 6×6 = 36
So, n(S)=6
Favorable cases where the sum is 10 or more with 5 on 1st die =(5,5),(5,6)
Event of getting the sum is 10 or more with 5 on 1st die = n(E) =2
Hence, the probability of getting a sum of 10 or more with 5 on 1st die
[tex] = \frac{n(E)}{ n(S)} [/tex]
[tex] = \frac{2}{36} [/tex]
[tex] = \frac{1}{18} [/tex]
[tex]\huge\bf\underbrace{\pink{A}\orange{n}\blue{s}\red{w}\green{e}\purple{r}}[/tex]
Solution : - x = 120°