SOLUTION: I think I have figured out this whole problem except for part D.
There are three boxes containing letters:
1: MATH
2: AND
3: HISTORY
A. From box 1, three letters are drawn
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-> SOLUTION: I think I have figured out this whole problem except for part D.
There are three boxes containing letters:
1: MATH
2: AND
3: HISTORY
A. From box 1, three letters are drawn
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Question 609421: I think I have figured out this whole problem except for part D.
There are three boxes containing letters:
1: MATH
2: AND
3: HISTORY
A. From box 1, three letters are drawn one-by-one without replacement and recorded in order. What is the probability that the outcome is HAT?
(1/4)(1/3)(1/2) = 1/24
B. From box 1, three letters are drawn one-by-one with replacement and recorded in order. What is the probability that the outcome is HAT?
(1/4)(1/4)(1/4) = 1/64
C. One letter is drawn at random from box 1, then another from box 2, and then another from box 3, with the results recorded in order. What is the probability that the outcome is HAT?
(1/4)(1/3)(1/7) = 1/84
D. If a box is chosen at random and then a letter is drawn at random from the box, what is the probability that the outcome is A?
In box 1 the Probability is 1/4, In box 2 the probability is 1/3, and in box 3 the probability is 0. I don't know exactly how to combine these to get an answer? Please let me know if I'm even close to getting the answers correct in this whole problem. Thanks so much! Answer by ewatrrr(24785) (Show Source):
Hi
three boxes containing letters: box 1: MATH , box 2: AND box 3: HISTORY
Super! You are Definitely on the right track! Yes! A. 1/24, B. 1/64, C. 1/84
D. If a box is chosen at random: box 1 'or' box 2 'or' box 3
Note: Probability of choosing a particular box = 1/3
P( A) = + + = = 21/108 = 7/36
