Question

The probability of 4 turning up at least once in two tosses of a fair die is \[\dfrac{{11}}{{{6^x}}}\], the value of x is?

Answer 1

Hint: Here we find the total possibilities in form of ordered pairs where first element is from the first toss and second element is from the second toss. Then favorable outcomes are the ones having one of the elements as 4 and both elements as 4. We use the method of probability to find the probability of 4 turning up at least once in two tosses of die and compare the value to the given value and write the value of x.
* Probability of an event is the number of favorable outcomes divided by total number of observations.
* A die has 6 faces each representing 1, 2, 3, 4, 5 and 6.

Complete step-by-step answer:
Since, we know a die has six outcomes. Therefore, outcomes from the first toss are 1, 2, 3, 4, 5 and 6 and outcomes from the second toss are 1, 2, 3, 4, 5 and 6.
So we make ordered pairs of outcomes such that the first element is from the first toss and second element is from the second toss. We can make a table of outcomes.

First toss\[ \to \]Second toss\[ \downarrow \]	1	2	3	4	5	6
1	(1,1)	(1,2)	(1,3)	(1,4)	(1,5)	(1,6)
2	(2,1)	(2,2)	(2,3)	(2,4)	(2,5)	(2,6)
3	(3,1)	(3,2)	(3,3)	(3,4)	(3,5)	(3,6)
4	(4,1)	(4,2)	(4,3)	(4,4)	(4,5)	(4,6)
5	(5,1)	(5,2)	(5,3)	(5,4)	(5,5)	(5,6)
6	(6,1)	(6,2)	(6,3)	(6,4)	(6,5)	(6,6)

So, from the table we see there are 6 columns and 6 rows.
Therefore, total number of observations is \[6 \times 6 = {6^2}\].
Now we find the favorable outcomes. We find the elements having at least one element as 4.
Elements having one of the element as 4 are: (1,4), (2,4), (3,4), (5,4), (6,4), (4,1), (4,2), (4,3), (5,3), (6,4).
Element having both elements as 4 are: (4,4)
Therefore, number of favorable outcomes are 11
Now we use the formula of Probability that is number of favorable outcomes divided by total number of outcomes.
Probability \[ = \dfrac{{11}}{{{6^2}}}\]
Since, we are given the probability as \[\dfrac{{11}}{{{6^x}}}\]
We compare the two values
\[ \Rightarrow \dfrac{{11}}{{{6^x}}} = \dfrac{{11}}{{{6^2}}}\]
Cross multiply the values on both sides and cancel the same terms.
\[{6^x} = {6^2}\]
So, we get \[x = 2\].

Note: Students many times make mistakes in writing the total observations if they write the values randomly, always try to make a table for observations. Also, many students ignore the possibilities where the elements are in opposite places of each other, keep in mind we take all the outcomes having four in any of the two elements.