Question

Suppose you like to keep a jar of change on your desk. Currently, the jar contains the following:

Answer 1

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

$\bf P(n|d)=\frac{P(d|n)P(n)}{P(d|p)P(p)+P(d|d)P(d)+P(d|n)P(n)+P(d|q)P(q)}$

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

$\bf P(n|d)=\frac{(24/73)(7/74)}{(24/73)(18/74)+(23/73)(24/74)+(24/73)(7/74)+(24/73)(25/74)}=0.0959$

So

P(n | d) = 0.0959