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Question: A five digit number of the form \({\text{x y z y x}}\) is chosen, probability that \({\text{x < y}}\...

A five digit number of the form x y z y x{\text{x y z y x}} is chosen, probability that x < y{\text{x < y}} is ?
A) 3590\dfrac{{35}}{{90}}
B) 615\dfrac{6}{{15}}
C) 1945\dfrac{{19}}{{45}}
D) 1330\dfrac{{13}}{{30}}

Explanation

Solution

Probability of an event can be found using favourable number of outcomes and total number of outcomes. We can choose digits for x,y,z{\text{x,y,z}} from zero to nine. But x cannot be zero. Thus we get the total number of cases. Also applying the condition x < y{\text{x < y}}, we get the favourable number of cases.

Formula used:
The probability of an event is the number of favourable outcomes divided by total number of outcomes.

Complete step-by-step answer:
Given that x y z y x{\text{x y z y x}} is a five digit number.
We have digits from one to nine to choose for x, y{\text{x, y}} and z{\text{z}} and also zero for y{\text{y}} and z{\text{z}}.
x{\text{x}} cannot be zero since the first digit cannot be zero.
So total number of possibilities for the number x y z y x{\text{x y z y x}} is 9×10×10=9009 \times 10 \times 10 = 900
The probability of an event is the number of favourable outcomes divided by total number of outcomes.
So let us find out the favourable number of outcomes.
We have the condition that x < y{\text{x < y}}.
So x{\text{x}} can be chosen from any of the digits one to nine and y{\text{y}} can be chosen from two to nine such that x < y{\text{x < y}}.
If we let x=1{\text{x}} = 1, then y{\text{y}} can be 2,3,4,5,6,7,8,92,3,4,5,6,7,8,9, 88 in total.
And if x=2{\text{x}} = 2, then y{\text{y}} can be 3,4,5,6,7,8,93,4,5,6,7,8,9.
Moving on, the possibilities get reduced by one.
Finally if x=8{\text{x}} = 8, then y{\text{y}} has only one possibility, which is 99
So total number of cases is 8+7+6+5+4+3+2+1=368 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 36
Now for each of these cases, z can be any digit from zero to nine.
That is corresponding to each of these 3636 cases, we have 1010 possibilities.
So the number of favourable cases is 36×10=36036 \times 10 = 360
Hence the probability is given by a favourable number divided by total number which is equal to 360900=615\dfrac{{360}}{{900}} = \dfrac{6}{{15}}.
Therefore the answer is option B.

Note: In the question it is not mentioned that the digits x, y{\text{x, y}} and z{\text{z}} are distinct. They can be the same as well. But in favourable outcomes x{\text{x}} and y{\text{y}} are distinct since one is strictly less than the other.