Question 1086272
Ray has some change in his pocket: four pennies, two nickels, four dimes,
two quarters, and two half dollars. He draws out a coin at random. The
probability of drawing any one of the coins is
0.06 for each penny.
<pre>
So since he has four pennies, his probability
of drawing a penny is 4×0.06 = 0.24
</pre>
0.01 for each dime.
<pre>
So since he has four dimes, his probability
of drawing a dime is 4×0.01 = 0.04
</pre>
0.03 for each nickel.
<pre>
So since he has two nickels, his probability
of drawing a nickel is 2×0.03 = 0.06
</pre>
0.09 for each quarter.
<pre>
So since he has two quarters, his probability
of drawing a quarter is 2×0.09 = 0.18
</pre>
0.24 for each half-dollar.
<pre>
So since he has two half-dollars, his probability
of drawing a half-dollar is 2×0.24 = 0.48

Now, let's see if the problem is legitimate by
adding up all those probabilities and seeing if
we get 1.00:

0.24 = probability of drawing a penny.
0.04 = probability of drawing a dime.
0.06 = probability of drawing a nickel.
0.18 = probability of drawing a quarter.
0.48 = probability of drawing a half-dollar.
----
1.00

Yes, we get 1, the probability of certainty, since
it is certain that he will get a penny or a dime or
a nickel or a quarter or a half-dollar, so this is 
a legitimate probability problem.
</pre>
(a) Find the probability that the coin drawn is a dime.
<pre>
0.04
</pre>
(b) Find the probability that the coin drawn is a quarter 
or a half-dollar.
<pre>
0.18 + 0.48 = 0.66 
</pre>
(c) Find the probability that the coin drawn is a nickel.
<pre>
0.06
</pre>
(d) Find the probability that the value of the coin drawn is 
less than 10 cents.   
<pre>
That's the same as if it had read this way: 

Find the probability that the coin drawn is a penny or a 
nickel.

0.24 + 0.06 = 0.30

Edwin</pre>