Drawer A contains five pennies and three dimes and drawer B

Drawer A contains five pennies and three dimes, and drawer B contains three pennies and seven dimes. A drawer is selected at random and a coin is selected at random from that drawer.

a) Find the probability of selecting a dime.

b) Suppose the dime is obtained. Find P(that the dime came from drawer B)

Solution

a) First of all, if a drawer was selected at random, both drawers have an equal chance

[P(A)=1/2 and P(B)=1/2]

You know that the probability of drawing a dime from Drawer A would be P(Dime given A)= 3/8 (or 3 dimes/8 coins total)

The probability of drawing a dime from Drawer B would be P(Dime given B)= 7/10 (or 7 dimes/10 coins total)

So the probability of drawing a dime =

P(Dime) = P(A) and P(Dime given A) or P(B) and P(Dime given B)

             =P(A) x P(Dime given A) + P(B) x P(Dime given B)

             =(1/2)(3/8) + (1/2)(7/10)

             = 0.1875   +   0.35

             = 0.5375

[Remember, in probability, \"and\" means multiply, and \"or\" means add]

b) Now in reverse, when a dime is obtained we simply find the probability of drawing a dime from B [P(B) and P(Dime given B)] GIVEN the probability that a dime is drawn (from part a ^ which = 0.5375

** \" X given Y\" means we know that Y happens for certain and we can divide out P(Y) if, and only if, it is accounted for in P(X)]

[P(B) and P(Dime given B)] GIVEN P(Dime)

= [(1/2) x (7/10)]/0.5375  

= 0.6512