Question 1131560
<br>
(a) P(at least one red ball) = 1-P(no red ball) = P(both balls blue)<br>
{{{(4/11)*(3/10) = 12/110 = 6/55}}}<br>
(b) P(second ball red | first ball red)<br>
By the definition of conditional probability, this is equal to<br><pre>
   P(first red AND second red)
  -----------------------------
        P(first red)</pre>
{{{((7/11)(6/10))/(7/11) = 6/10 = 3/5}}}<br>
I, as well as many students, find the formal definition of conditional probability, in the form of a formula for calculating it, somewhat confusing.  For me it is easier to understand this way:<br>
"...given that the first ball is red" means the only cases we are considering are those in which the first ball is red; that means the denominator of the probability fraction, representing the "sample space", is only the probability of getting red with the first ball: P(first red).  Then the numerator is the probability that we ALSO get a red on the second ball: P(first red AND second red).<br>
(c) P(second ball red | first ball blue)<br>
This is<br><pre>
   P(first blue AND second red)
  -----------------------------
        P(first blue)</pre>
{{{((4/11)(7/10))/(4/11) = 7/10}}}