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A bag contains twenty Rs \[\mathbf{5}\] coins, fifty Rs \[\mathbf{2}\]coins and thirty Re \[\mathbf{1}\] coins. If it is equally likely that one of the coins will fall down when the bag is turned upside down, what is the probability that the coin: \[\left( \mathbf{iii} \right)\]will neither be a Rs \[\mathbf{5}\] coin nor be a Re \[\mathbf{1}\] coin?
A bag contains twenty Rs \[\mathbf{5}\] coins, fifty Rs \[\mathbf{2}\]coins and thirty Re \[\mathbf{1}\] coins. If it is equally likely that one of the coins will fall down when the bag is turned upside down, what is the probability that the coin: \[\left( \mathbf{iii} \right)\]will neither be a Rs \[\mathbf{5}\] coin nor be a Re \[\mathbf{1}\] coin?
CBSE | Class 10 | Exercise 13.3 | Selina | Statistics and Probability
A bag contains twenty Rs \[\mathbf{5}\] coins, fifty Rs \[\mathbf{2}\]coins and thirty Re \[\mathbf{1}\] coins. If it is equally likely that one of the coins will fall down when the bag is turned upside down, what is the probability that the coin: \[\left( \mathbf{iii} \right)\]will neither be a Rs \[\mathbf{5}\] coin nor be a Re \[\mathbf{1}\] coin?

Solution:

\[\left( iii \right)\] Number of favourable outcomes for neither Re \[1\]nor Rs \[5\]coins \[=\] Number of favourable outcomes for Rs\[~2\] coins \[=\text{ }50\text{ }=\text{ }n\left( E \right)\]

Hence, probability \[\left( neither\text{ }Re\text{ }1\text{ }nor\text{ }Rs\text{ }5\text{ }coin \right)\text{ }=~n\left( E \right)/\text{ }n\left( S \right)\text{ }=\text{ }50/100\text{ }=\text{ }1/2\]

i

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