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a box contains 6 quarters and 4 dimes. three of the coins are randomly selected and it is known that they are not all dimes. what is probability that both dimes and quarters were selected

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The probability of selecting a coin from the jar that is not a dime or a penny is 10/23.

To determine the probability of selecting a coin from the jar that is not a dime or a penny, first calculate the total number of coins in the jar:

Total coins = Quarters + Dimes + Nickels + Pennies

Total coins = 6 + 3 + 4 + 10

Total coins = 23

Next, calculate the probability of selecting a dime or a

Probability (Dime or Penny) = (Dimes + Pennies) / Total coins

Probability (Dime or Penny) = (3 + 10) / 23

Probability (Dime or Penny) = 13/23

Finally, subtract this probability from 1 to find the probability of not selecting a dime or a penny:

Probability (Not Dime or Penny) = 1 - Probability (Dime or Penny)

Probability (Not Dime or Penny) = 1 - 13/23

Probability (Not Dime or Penny) = 10/23

Therefore, the probability of selecting a coin from the jar that is not a dime or a penny is 10/23.

Question

jar contains 6 quarters, 3 dimes, 4 nickels, and 10 pennies. if a coin is randomly selected from the jar, what is the probability that it will not be a dime or a penny?

User Qollin
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