Probability

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mollymvora

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Feb 21, 2011
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The problem says, "When Carlos shoots a basketball, the probability of a basket is 0.4. When Brad shoots, the probability of a basket is 0.45. What is the probability that at least one basket is made if they each take one shot.
P(both) = .45 * .4
P(Carlos) = .4 * .55
P(Brad) = .6 * .45

Then I add them. However, what i do not understand is that why don't we multiply what P(Carlos) and P(Brad) by two because it could be Carlos Makes, Brad does not make OR Brad does not make, Carlos Makes. Which means that it is two different possibilities. Does the order not matter in this problem?
 
Maybe this way of doing will help. When you see "at least one", try finding the probability of none and subtracting from 1.

In this case, the probability Carlos does not make the shot is .6

The probability Brad does not make the shot is .55

So, \(\displaystyle 1-(.55)(.6)=.67\). Same as your result.

If it were \(\displaystyle 1-(.6)(.55)\) it is still the same result.
 
mollymvora said:
The problem says, "When Carlos shoots a basketball, the probability of a basket is 0.4. When Brad shoots, the probability of a basket is 0.45. What is the probability that at least one basket is made if they each take one shot.
P(both) = .45 * .4
P(Carlos) = .4 * .55
P(Brad) = .6 * .45

Then I add them. However, what i do not understand is that why don't we multiply what P(Carlos) and P(Brad) by two because it could be Carlos Makes, Brad does not make OR Brad does not make, Carlos Makes. Which means that it is two different possibilities. Does the order not matter in this problem?
Click to expand...

Order does not make a difference because whether Carlos makes the shot (or not) - does NOT depend on - whether Brad makes the shot (or not).
 
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