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04 Oct 2024, 00:13
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C
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Difficulty:
95%
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Question Stats:
48% (03:04) correct
52%
(03:02)
wrong
based on 69
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A basket contains exactly 16 fruits, 10 of which are apples and the rest are oranges. In the bar graph shown above, the middle and the right bars are the same height and the left bar is one-third this height. This height is intended to represent, respectively, the probabilities of obtaining 0 apples, exactly 1 apple, and exactly 2 apples when 2 fruits are randomly selected without replacement from the basket. However, 1 of the bar heights is incorrect. The correct height is x times the height shown.
Which of the following choices for the bar and the value of x would result in an accurate representation of the labelled probabilities?
A. Left bar, x = 4/3
B. Middle bar, x = 3/4
C. Middle bar, x = 4/3
D. Right bar, x = 3/4
E. Right bar, x = 4/3
04 Oct 2024, 00:18
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04 Oct 2024, 00:30
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MT1302
- P(0 apples out of 2) = 6/16 * 5/15 = 1/8
- P(1 apples out of 2) = 6/16 * 10/15 * 2 = 1/2 = 4/8
- P(2 apples out of 2) = 10/16 * 9/15 = 3/8
The third result is three times the first one, so bars 1 and 3 are consistent with the given information and are correct. Therefore, the height of the second bar is not correct. The second bar shows 3/8 (since we are told that the heights of the second and third bars are equal), but it should actually represent 4/8. Thus, the second bar needs to be adjusted to (4/8) / (3/8) = 4/3 times its current height.
Answer: C.
