laythesmack23 wrote:
Jay Z is rolling a number cube with faces numbered 1 to 6 repeatedly. When he receives a 4, he will stop rolling the cube. What is the probability that Jay Z will roll the die less than 3 times before stopping?
Please explain in detail, I'm having problems setting the problem up. If there are multiple ways of setting the problem up, please list those possibilities, thanks!
Hello! I'm happy to help with this.
This is a relatively challenging probability question --- this is typical, for example, of a question on probability theory that the AP Statistics exam would ask. This is getting toward the outer limit of anything the GMAT would ask about probability.
First, let's think about an individual trial ---- one roll of the die. We are going to call getting a 4 a "success." On one trial,
P(
success) = 1/6
and
P(
not success) = 5/6
Now, we have to figure out the probability of getting a successful trial in fewer than three trials.
(BTW --- grammar point important for GMAT SC -- it should be
fewer than three trials, not
less than three trials, because trials are countable. See:
http://magoosh.com/gmat/2012/gmat-gramm ... -vs-fewer/)
Let N = the number of the trial on which the first successful trial occurs.
The restriction "fewer than three trials" means we are looking at the cases N = 1 or N = 2 only.
Case #1: N = 1
In this scenario, we toss the cube once and it's successful. Probability =
1/6Case #2: N = 2
This scenario involves two tosses, and has the requirements (a) the first toss is not a success, and (b) the second toss is a success. Because those two are joined by the word "and", that means
multiply in probability.
P(1st toss =
not success and 2nd toss =
success) = (5/6)*(1/6) =
5/36Calculating the total probability involves an "or" statement --- in probability, "or" means
add.
P(N<3) = P(N= 1
or N = 2) = 1/6 + 5/36 =
11/36Does all this make sense?
Mike
_________________
Mike McGarry
Magoosh Test Prep