Question

You reach into a bag of coins and withdraw two coins without replacement. What is the probability you withdrew a nickel and then a dime if the bag held five pennies, ten nickels, and four dimes?

Answer 1

20/171







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Answer 2

Answer:
`(20)/(171)`

Explanation:

Given : Number of nickels in the bag = 10

Number of dimes in the bag = 4

The total number of coins in the bag =
`10+4+5=19`

The probability of drawing nickel :-

`\text{P(nickel)}=(10)/(19)`

Total number coins left = 18

The conditional probability of drawing dime, given that a nickel is already drawn :-

`\text{P(dime}|\text{nickel)}=(4)/(18)`

Now, the probability you withdrew a nickel and then a dime is given by :-

`\text{P(dime and nickel)}=\text{P(dime}|\text{nickel)}*\text{P(nickel)}\\\\=(10)/(19)*(4)/(18)=(20)/(171)`

Now, the probability you withdrew a nickel and then a dime
`}=(20)/(171)`