A class with 10 kids lines up for recess. Two of the kids in the class are named Lucy and Frank.

Question

A class with 10 kids lines up for recess. Two of the kids in the class are named Lucy and Frank. Assume that the order in which the kids line up is random (that is, all outcomes are equally likely). What is the probability that Lucy is first in line or Frank is last in line?

$$\frac{\binom{10}{9}}{2^{10}}+\frac{\binom{10}{9}}{2^{10}}=\frac{5}{2^8}$$

Can anyone verify this answer please..

Answer 1

The probability that Lucy is first is $\frac 1{10}$. The probability that Frank is last is $\frac 1{10}$. The probability that both happens is $\frac1{10}\cdot \frac 19$. So the probaility of "OR" is $\frac1{10}+\frac1{10}-\frac1{90}=\frac{17}{90}$

Answer 2

Each of the terms in your sum is $1/2^{10}$. You seem to think one is the probability that Lucy is first, the other the probability that Frank is last. But they are much too small for that.

Their sum is not $1/2^8$.

Moreover, you haven't thought through the problem. Solve it for a class with just two or three kids by listing all the possibilities. Then generalize.

Answer 3

• There are $10!$ arrangements altogether.
• There are $9!$ arrangements with Lucy first.
• There are $9!$ arrangements with Frank last.
• There are $8!$ arrangements with Lucy first and Frank last. So, all together $$\frac{9!+9!-8!}{10!}= \frac{8!}{10!}(2\cdot 9 - 1)=\frac{17}{90}$$