Question

Find the probability distribution of the number of success in two tosses of die, where a success is defined as:
Number is greater than 4.
Six appears on at least one die.

Answer 1

Hint: First write down all the possible outcomes when two dice are tossed. Then for each of the cases consider all the possible ways for an event to happen. In case-1, consider all the numbers, which are greater than 4. In case-2, find out all the outcomes from the total outcomes which have at least one 6 in them. Then find out the probabilities for each case one by one.

Complete step-by-step answer:

A probability distribution is a statistical function that describes all the possible values and likelihoods that a random variable can take within a given range. This range will be bounded between the minimum and the maximum possible values.
We know that a die has six faces with six different values starting from 1 to 6.
 When a die is tossed two times we can get $6\times 6=36$ number of observations.
Let X be a random variable, which represents the number of successes.
A random variable is a numerical description of the outcome of a statistical experiment.
For the first case the success refers to the number greater than 4.
When we throw to dies, there can be three cases:
First one is that no number is greater than four.
Second one is that only one number is greater than four.
Third one is both the numbers are greater than four.
Let the probability of getting less than or equal to 4 on both the tosses is denoted by:
$P\left( X=0 \right)$
For this case the favourable number of outcomes are 4, and the total number of outcomes are 6.
Hence, $P\left( X=0 \right)=\dfrac{4}{6}\times \dfrac{4}{6}=\dfrac{4}{9}$
Let the probability of getting less than or equal to 4 on the first toss and greater than 4 on the second toss + number greater than 4 on first toss and less than or equal to 4 on the second toss is denoted by:
$P\left( X=1 \right)$
Hence, $P\left( X=1 \right)=\dfrac{4}{6}\times \dfrac{2}{6}+\dfrac{2}{6}\times \dfrac{4}{6}=\dfrac{4}{9}$
Let the probability of number greater than 4 on both the tosses is denoted by:
$P\left( X=2 \right)$
Hence, $P\left( X=2 \right)=\dfrac{2}{6}\times \dfrac{2}{6}=\dfrac{1}{9}$
Therefore our probability distribution function is:

$X$	0	1	2
$P\left( X \right)$	$\dfrac{4}{9}$	$\dfrac{4}{9}$	$\dfrac{1}{9}$

For the second case success means six appears on at least one die.
Here we can get two cases.
First one is six and does not appear on any of the tosses.
Second one is six appears on at least one of the tosses.
Let the success be defined by the random variable Y.
Let the probability of appearing six at least on one of the tosses is denoted by:
$P\left( Y=1 \right)$
Therefore, $P\left( Y=1 \right)\dfrac{5}{6}\times \dfrac{1}{6}+\dfrac{1}{6}\times \dfrac{5}{6}+\dfrac{1}{6}\times \dfrac{1}{6}=\dfrac{11}{36}$
Let the probability of six does not appear on any of the tosses is denoted by:
$P\left( Y=0 \right)$
Therefore, $P\left( Y=0 \right)=1-P\left( Y=1 \right)=1-\dfrac{11}{36}=\dfrac{25}{36}$
Hence the probability distribution function is:

$X$	0	1
$P\left( X \right)$	$\dfrac{25}{36}$	$\dfrac{11}{36}$

Note: We generally make mistakes while considering the number of possible cases for an event to happen. We have to take all the possible cases otherwise the distribution function will be wrong.
There is a good way to check if the distribution function is correct or not. If we add all the probabilities you should always get one. Otherwise our answer is wrong.
For example for the first case: $\dfrac{4}{9}+\dfrac{4}{9}+\dfrac{1}{9}=\dfrac{4+4+1}{9}=\dfrac{9}{9}=1$