When 2 dice are thrown simultaneously what is the probability that there is exactly one 5?
A) \[\dfrac{4}{{36}}\]
B) \[\dfrac{5}{{18}}\]
C) \[\dfrac{6}{{23}}\]
D) \[\dfrac{7}{{24}}\]

Hint: Both the dices can have the same set of outcomes i.e., 1,2,3,4,5,6 . Now everything depends upon the outcome where only one will be 5 and the other can be anything from 1 to 6 other than 5.

Complete step by step solution:
So we will only take the solution where 5 occurs exactly once. So as for one dice to have 6 possible outcomes it can be easily noted that 2 dice thrown simultaneously will give 36 outcomes and from there we need to choose, how many of them will have the outcomes required by us.
So let us try to do it by writing what the favourable outcome could look like, so if we consider the values to be represented as (a,b) where a and b both can take any value from 1 to 6. So for our favourable outcomes exactly one 5 should appear once. So let us fix the value of a as 5 and then the value of b. Thus the outcomes becomes (5,1),(5,2),(5,3),(5,4),(5,5),(5,6) and similarly we can fix the value of 5 in b and change the values of a so so the outcomes will be (1,5),(2,5),(3,5),(4,5),(5,5),(6,5)
Now it was known to us that exactly one 5 can be there therefore (5,5) and (5,5) from both the sets will be excluded which means No. of favourable outcome is \[5 + 5 = 10\] and the total number of outcomes is 36.
Therefore the probability is the total number of favourable outcome divided by the total number of possible outcome
\[\therefore p = \dfrac{{10}}{{36}} = \dfrac{5}{{18}}\]
Which means that B is the correct option.

Note: Another approach for this question can be Consider dice-1 gives you 5. The probability of getting 5 is \[\dfrac{1}{6}\] (as there are 6 possible cases and you are choosing 1). Now the dice-2 should give us any value except 5. So the probability of getting the desired result is \[\dfrac{5}{6}\] (as we can choose 5 possible values out of 6).
So the total probability of getting 5 only in dice-1 and all other possible values in dice-2 is \[\dfrac{1}{6} \times \dfrac{5}{6} = \dfrac{5}{{36}}\] But we can also think of a situation where dice-2 gets 5 and dice-1 gets all the possible outcomes. And the probability of this will also be \[\dfrac{1}{6} \times \dfrac{5}{6} = \dfrac{5}{{36}}\] Finally, the total probability of getting 5 exactly once when two dice are thrown in \[\therefore p = \dfrac{5}{{36}} + \dfrac{5}{{36}} = \dfrac{{10}}{{36}} = \dfrac{5}{{18}}\] .