Question
Question: A pair of dice is tossed. How do you find the probability of the sum of the numbers being \[8?\]...
A pair of dice is tossed. How do you find the probability of the sum of the numbers being 8?
Solution
In the probability types problems, first we try to obtain the total outcomes and then we try to find out the favourable outcomes because by definition of probability we know that probability of an event is the ratio of a favourable outcome to the total number of outcomes. So, after obtaining both the values we divide the favourable outcome by the total number of outcomes so that the desired value can be obtained. Now, we can proceed to solve the given problem.
Formula used:
Probability of an event,P(A) = $$$$\dfrac{{n(A)}}{{n(S)}}
Where, n(A)= number of favourable outcome and n(S)= number of total outcomes
Complete step-by-step solution:
First try to find out the total number of possible outcomes, when a pair of dice is tossed the following outcomes are possible,
⇒(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)
⇒(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)
⇒(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)
⇒(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)
⇒(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)
⇒(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)
Therefore, the number of total outcomes becomes equal to 36.
It can also be written as following,
⇒n(S)=36
Now taking those outcomes from the above all outcomes that when added gives 8. This will become the favourable outcomes,
⇒(2,6)(6,2)(5,3)(3,5)(4,4)
As when we add all above outcomes it gives 8 as following,
⇒(2+6)=8
⇒(6+2)=8
⇒(5+3)=8
⇒(3+5)=8
⇒(4+4)=8
Therefore, the number of favourable outcomes becomes equal to 5.
It can also be written as following,
⇒n(A)=5
Now applying above given formula of probability, we can write it as following,
\Rightarrow $$$$P(A)$$$$ = $$$$\dfrac{{n(A)}}{{n(S)}}
Putting the values in above formula, we get
\Rightarrow $$$$P(A)$$$$ = $$$$\dfrac{5}{{36}}
Therefore, we got the required probability which is equal to 365.
Note: Probability of any event occurs between ‘0’ and ‘1’ where ‘0’ can be said as an impossible event and ‘1’ can be said as a sure event. the probability depends on the favourable outcomes.