Science:Math Exam Resources/Courses/MATH105/April 2012/Question 08 (e)
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Question 08 (e) |
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Short-answer questions. No credit will be given for the answer (even if it is correct) without the accompanying work. There are two boxes containing two balls each. The balls in the first box are numbered −1 and 1, the balls in the second box are numbered 0 and 2. An experimenter draws a ball from each box and observes the number of each ball. Define a random variable X whose value is two times the number of the ball drawn from the first box plus three times the number of the ball drawn from the second box. In other words, X = 2(number observed from box 1) + 3(number observed from box 2). Write down all possible values of X and use this to compute the expected value of X. |
Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you? |
If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! |
Hint |
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There are four possible outcomes for the experiment and hence X can take four possible values. Try listing them out. All outcomes are equally likely. |
Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work.
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Solution |
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Please rate my easiness! It's quick and helps everyone guide their studies. We consider all possible outcomes. Draw -1 from bin 1 and 0 from bin 2: . Draw -1 from bin 1 and 2 from bin 2: . Draw +1 from bin 1 and 0 from bin 2: . Draw +1 from bin 1 and 2 from bin 2: . All possible values of X are: -2, 2, 4, 8. Each outcome is equally likely since drawing from the bins is independent and within each bin the balls have equal probability (this isn't mentioned, but it seems like a reasonable assumption). Therefore, (since the sum of all 4 probabilities must be 1). The expected value of X is: |