Question

 A die is thrown and the sum of the numbers appearing is observed to be 6. What is the probability that the number 4 has appeared at least once?

Answer 1

Hint: Take or consider two events A and B where A is number out come for which sum is 6 and B is number of outcomes for which a number 4 is there. Then find the common number of outcomes of A and B. So the probability will be \[\dfrac{\text{Number of common outcomes of B and A}}{\text{Number of outcomes of A}}\].

Complete Step-by-Step solution:
In this question we are said that a die is thrown twice and the sum of numbers appearing is observed to be 6. So we have to find the probability that the number 4 has appeared once.
We will find the probability using formula, \[P\left( \dfrac{B}{A} \right)=\dfrac{n\left( A\cap B \right)}{n\left( A \right)}\]
Here \[P\left( \dfrac{B}{A} \right)\] Probability of getting event B considering A has already happened. \[n\left( A\cap B \right)\]Means number of which is common to both A and B. \[n\left( A \right)\]means number of outcomes favorable to event A irrespective of what B is.
Now let’s consider event A as the number of outcomes for which the sum of the number in two die is 6 and event B as the number of outcomes for which number 4 appeared once at least.
Let’s write the outcomes if two dies is thrown:
{(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)}
So here are the total 36 outcomes.
For the event A the outcomes will be: (1, 5), (2, 4), (3, 3), (4, 2), (5, 1)
For the event B the outcomes will be: (1, 4), (2, 4), (3, 4), (4, 4), (5, 4), (6, 4), (4, 1), (4, 2), (4, 3), (4, 5), (4, 6)
\[n\left( A\cap B \right)\] will be 2 as only (4, 2), (2, 4) outcomes satisfy.
\[n\left( A \right)\] will be 5.
So, required Probability is = \[\dfrac{2}{5}\]

Note: We can directly do the solution by first considering only those outcomes whose sum is 6 which are \[\left( 1,5 \right)\] \[\left( 2,4 \right)\]\[\left( 3,3 \right)\] \[\left( 4,2 \right)\] \[\left( 5,1 \right)\] from which only \[\left( 2,4 \right)\] and \[\left( 4,2 \right)\] have a four in each. So we can say the probability is \[\dfrac{2}{5}\]