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Two different dice are tossed together. Find the probability:
(i) of getting a doublet
(ii) of getting a sum 10, of the number on the two dice.

Answer
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Hint: At first write down the total number of outcomes then find favorable events of each case then use the formula to find the desired result.

Complete step-by-step answer:
At first, we will define what is probability and understand the basic terms related to the probability to be used in the question.
The probability of an event is a measure of the likelihood that the event would occur.
If an experiment’s outcome is equally likely to occur, then the probability of an event E is the number of outcomes in E divided by number outcomes in the sample space. Here sample space consists of all the events that can occur possibly.
So, it can be written as $P(E)=\dfrac{n(E)}{n(S)}$
Here p(E) is probability of an event or events which is asked, n(E) is number of favorable events and n(S) is number of all the events that can occur possibly.
(i) So here we are asked to find the probability of getting a doublet
So, the sample space consists of 36 elements
So, n(S) = 36
Now for favorable events, only pairs are (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6) which means only 6 elements.
So, n(E) = 6
So, the probability is, $\begin{align}
  & P(E)=\dfrac{n(E)}{n(S)} \\
 & =\dfrac{6}{36} \\
 & =\dfrac{1}{6}
\end{align}$
Hence the probability is $\dfrac{1}{6}$ .
(ii) So here we are asked to find the probability of getting sum 10 on the dice.
So, the sample space consists of 36 elements
So, n(S) = 36
Now for favorable events only pairs (4, 6), (5, 5), (6, 4) satisfy
So, n(E) = 3
So, the probability is $\begin{align}
  & P(E)=\dfrac{n(E)}{n(S)} \\
 & P(E)=\dfrac{3}{36}=\dfrac{1}{12} \\
\end{align}$
Hence the probability is $\dfrac{1}{12}$
The probability is $\dfrac{1}{6}$ and $\dfrac{1}{12}$respectively.
Note: Students should be careful while counting the total events and also total number of favorable events. They should also be careful about using formulas of probability.