X and Y plays a game in which they are asked to select a number from $21 - 50$ . If the two number match both of them wins a prize. Find the probability that they will not win a prize in the single trial.

\frac{1}{5}

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\frac{21}{84}

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\frac{28}{30}

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\frac{29}{30}

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Solution

The correct option is D $\frac{29}{30}$
Total numbers $= 30$
$\Rightarrow$ Probability that both of them gets same number $= \frac{1}{30}$
$\Rightarrow$ Probability that they get different numbers $= \frac{29}{30}$
$∴$ Probability that they will not win $=$ probability that they get diffetenr numbers $= \frac{29}{30}$
Hence, the answer is $\frac{29}{30} .$

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X and Y plays a game in which they are asked to select a number from 21−50. If the two number match both of them wins a prize. Find the probability that they will not win a prize in the single trial.

The correct option is D 2930Total numbers =30⇒ Probability that both of them gets same number =130⇒ Probability that they get different numbers =2930∴ Probability that they will not win = probability that they get diffetenr numbers =2930Hence, the answer is 2930.

X and Y plays a game in which they are asked to select a number from $21 - 50$ . If the two number match both of them wins a prize. Find the probability that they will not win a prize in the single trial.