AboutJason Eisele Expertise I am qualified to answer probability questions through the undergraduate level. I can also assist with the first actuarial exam in probability and explain the roles of probability in applications such as economics and game theory.
I hope to address any topics that I am currently unfamiliar with during my studies as an actuary.
Experience I am beginning employment as an Actuarial Assistant with a major auto insurer. On the side, I have experience applying probability to poker at a highly competitive level.
Organizations Mensa
Education/Credentials University of Rochester Class of 2008:
Bachelor of Arts, Mathematics and Economics
Certificate in Actuarial Studies
Certificate in Mathematical Modeling in Political Science and Economics
Certificate in Management Studies with track in Accounting and Finance
Question A softball player has a 30% chance of getting a hit each time she comes to bat. Assuming that each at-bay is a statistically independetn event, what is the probability that she will get a hit in her first at bat?
What is the probability that she will get a hit in her first or her second at bat?
What is the probability that she will get foru hits in a row?
To win a lottery where she has to match 5 numbers drawn at random, with the numbers ranging from 1 to 35. Once a number is drawn, it is not replaced. She does not have to match teh number in order; they just have to matich. What is her probability of winning?
Answer Hi Emily,
These are good questions that test your fundamental knowledge of probability concepts. Hopefully reviewing these can help you solve similar problems in the future!
I'll assist with the softball question first:
-The probability of getting a hit in her first at bat is given to you in percentage form as 30%. All you have to do is convert this to decimal form by taking the percentage sign away and moving the decimal point to the left two places, making it .30. This can be simplified to .3, or given in three-place batting average form as .300.
-We can use the following formula to determine the probability of either of two events happening:
P(A or B) = P(A) + P(B) - P(A and B)
In plain English, you can find the probability of either of two events occurring by adding the probabilities of each event occurring individually, and then subtracting from that the likelihood of both happening. This is because we cannot double-count the circumstance in which two successes (hits) occur. In this case P(at least one hit in two at bats) = .3 + .3 - (.3*.3) = .6 - .09 = .51
Another way to solve a problem like this is to find the probability of no hits in two at bats and then subtract that answer from 1. We can to this because she will either get no hits or at least one hit (these outcomes are exhaustive) but she cannot possibly get no hits and at least one hit at the same time (these outcomes are mutually exclusive). When we have exhaustive and mutually exclusive events, their probabilities must add up to 1. So in this case, the probability if getting no hits is .7 * .7 = .49, and thus the probability of getting at least one hit is 1 - .49 = .51.
-If you can answer the previous part, you can answer this one. We multiply just like before: .3^4 = .3 * .3 * .3 * .3 = .0081
And now for the lottery question:
The first step is to figure out how many ordered combinations of 5 numbers we can have from these 35 choices. She can pick any of 35 numbers first, then any of 34 second, any of 33 third, any of 32 fourth, and any of 31 fifth. Thus we have 35*34*33*32*31 possible ordered sets. However there are 5! = 5*4*3*2*1 ways to order a collection of 5 numbers (by the same logic as above), so we can divide 35*34*33*32*31 by 5*4*3*2*1 to find the number of possible unordered sets.
Because this is such a common form for a problem, there is a formula for finding the number of possible combinations:
n choose r = n!/[r!(n-r)!]
35 choose 5 = 35!/(5!30!)
You'll notice that (n-r)! eliminates all but the top r factors in n!. Thus if you write all of the factors that aren't both on the top and bottom, you should have r numbers left in each (just like the 5 each above).
n choose r is typically written as an 'n' above an 'r', both in the same parentheses, and it can be found on a calculator as nCr.
Once we know that there are 35 choose 5 = 324,632 unordered combinations and that there is only one that she can match, we know that she has a 1/324,632 probability of winning.
I hope this helps you solve more problems like this!
