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A and B play a game where each is asked to select a number from 1 to 25. If the two numbers match, both of them win a prize. The probability that they will not win a prize in a single trial is
(a)$\dfrac{1}{25}$
(b)$\dfrac{24}{25}$
(c)$\dfrac{2}{25}$
(d)$\dfrac{1}{625}$

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Hint: Firstly, we will find the total number of the ways in which numbers can be selected by A and B. Then, we will find the number of the ways in which either player can choose the same answers. Then, we will find the probability to win the game in a single trail. Then, subtract the probability to win the game in a single trail from the total probability which is 1 to get the probability that the game is not won in the single trail.

Complete step by step solution:
In this question, we are supposed to find the total number of the ways in which numbers can be chosen by A and B.
The total number is 25 and both have the access of choosing all numbers.
So, the total number of ways in which numbers are chosen is
$25\times 25=625$
Now, we will find the number of ways in which either player can choose the same numbers which is 25.
So, to find the probability of winning the prize which is number chosen is same is given by the ratio of the favourable cases to the total cases:
$\dfrac{25}{625}=\dfrac{1}{25}$
Now, to get the probability that A and B will not choose the same number and don't win the prize is given by the subtraction of the probability to win the game in a single trail from the total probability which is 1.
$\begin{align}
  & 1-\dfrac{1}{25} \\
 & \Rightarrow \dfrac{25-1}{25} \\
 & \Rightarrow \dfrac{24}{25} \\
\end{align}$

Note: The common mistakes done by you in this type of the question is that you will forget to take the total number of cases as 625 and consider it as 25 due to the fact that the number range given is 1 to 25. But it is clearly mentioned in the question that both A and B will select the number from that range which gives the total outcomes as $25\times 25=625$.
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