Answer:
0.0959
Step-by-step explanation:
There are 18+24+7+25=74 coins in the jar
Let's call
P(p) = 18/74 the probability of grabbing a penny
P(d) = 24/74 the probability of grabbing a dime
P(n) = 7/74 the probability of grabbing a nickel
P(q) = 25/74 the probability of grabbing a penny
What is the probability that you reach into the jar and randomly grab a dime and then, without replacement, a nickel?
Here we want to find P(n | d) the probability of grabbing a nickel given that you already grabbed a dime.
By the Bayes' Theorem
Now,
P(d | p) = 24/73 since there are now 73 coins and 24 dimes.
Similarly,
P(d | d) = 23/73 for you already grabbed a dime
P(d | n) = 24/73
P(d | q) =24/73
Replacing in the Bayes' formula
So
P(n | d) = 0.0959