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Tough Interview Questions...and Answers

Dr. B,

The interviews went well this past week. I should be hearing back from the both this coming week. But here is something interesting you may share with your classes. During my interviews with Veritas for a geophysicist position, I was asked several math and physics questions. Here they are in the order they were asked.

1) You have a bucket with 8 yellow, 2 red, and 3 blue balls. What is the probability that you draw a blue ball, then another blue ball, then a yellow ball?

2) You have that same bucket. What is the probability that you draw 2 blue balls and a yellow ball when selecting 3 at once? Compare this probability with the first one and if different, explain why.

3) You have a cannon on flat ground that is aimed 30 degrees up from the horizontal. Neglecting air resistance, how far will the cannon fly if it is fired at 100 feet/sec?

4) Derive the trigonometric identity: sin^2(x) + cos^2(x) = 1

5) (This question was asked by the VP when it was just he and I in the room). What is the probability that we have the same birthday (month and day)?

6) If you have 40 people in a room, what is the probability that at least 2 have the same birthday (month and day)?

7) You have a mass M hanging from a massless string of length L and you pull the mass back at an angle X. If you let it go, where does the maximum velocity occur and what is the maximum velocity?

8) You have a box with an object in it. You want to find out what the shape of the object is so you take 3 xrays of the box; one from the front, one from the side, and one from the top. One xray shows a triangle, one shows a circle, and the other shows a shape that looks like the silhouette of a propane tank (two flat sides and two rounded "sides"). Describe the object in the box.

These questions weren't all asked at once or all by the same person. They were randomly asked at different times. Also, several times they tried to disrupt my train of thought when I was working through a problem by asking me questions like what time it was, what the date was, where I lived, what kind of car I drive, etc. They were very unique interviews. But i think I did a pretty good job. I'll let you know if I hear anything this week. Let me know if you want me to email you my solutions.

Take Care
Brandt

Here are Brandt's answers:

1) Since you are drawing one ball on three separate events, you just look at the probability of each event and then multiply them all together. For a blue, then another blue, then a yellow: (3/13)*(2/12)*(8/11)

2) This one is different because you are drawing 3 at a time and there is no specific order. Since you already calculated the probability of one arrangement of two blues and one yellow in the first problem, simply multiply that answer by the number of distinct times you can arrange two blues and one yellow. blue blue yellow, blue yellow blue, yellow blue blue. So multiply the answer in problem one by 3

3) This one was pretty straight forward and easy.

4) This one was easy as well. I just used a circle of radius R and Pythagorean's Theorem.

5) There are 365 possible birthdays that two people can share. There are 365^2 different combinations of two birthdays. The solution is simply the ratio of (365)/(365^2), or 1/365.

6) This one requires the use of some probability theory. The sum of the probabilities of every possible event will equal 1. The different events for this problem are: no one having the same birthday, 2 people having the same birthday, 3 people having the same birthday, 4 people having the same birthday, 5 people having the same birthday, . . . , 40 people having the same birthday. The event of at least 2 people having the same birthday includes 2, 3, 4, 5, . . ., 40 people. This event only leaves out the event of no one having the same birthday. So calculating the probability of no one having the same birthday, and subtracting that from one, will give us the probability that at least 2 people have the same birthday. To calculate none having the same birthday, we must select 40 different birthdays from 365. This is done by using the combinations rule: 365 C 40, or mathematically: (365!)/(325!*40!). This is the total number of ways we can get 40 different birthdays out of the total 365 available. The total number of ways to get 40 birthdays from 365 without any restriction on them all being different is 365^40. So the probability of no one in a room of 40 having the same birthday is the ratio (365 C 40)/(365^40). Now subtracting this ration from 1 will give the solution to the problem.

7) This one is fairly straightforward

8) This one is difficult to describe but easy to picture. Take a cylinder about the size of a coke can. Lay it down so that you see the side, not the top or bottom. From the center of the can at the very top, place a knife so that it is parallel with the sides of the can (the sides are really the top and bottom of the can if it was standing upright). Now cut down to the very bottom left part of the can and then again from the top center to the bottom right part of the can. So staring at it from the side you should see a triangle. If you look at it up top, you see something that resembles a hotdog. The last view will give you a circle. I had some trouble with this problem in the interview.

Brandt

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