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The probability that Frances gets a line busy whenever she calls to
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12 Nov 2024, 00:23
12 Nov 2024, 00:23
Expert Reply
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Question Stats:
71% (01:10) correct
28% (02:13) wrong based on 7 sessions
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The probability that Frances gets a line busy whenever she calls to Michael is 2/3. She makes one call to Michael on each of the 4 consecutive days.
Quantity A |
Quantity B |
Probability that Frances gets the line busy on each of the 4 days |
1/4 |
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The probability that Frances gets a line busy whenever she calls to
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15 Nov 2024, 10:59
15 Nov 2024, 10:59
1
This is an example of an independent event, so the total probability of Frances getting a busy line on all 4 days is the product of the probabilities of her getting a busy line each day.
P(busy line on day 1 AND day 2 AND day 3 AND day 4) = P(busy line on the 1st day) * P(busy line on the 2nd day) * P(busy line on the 3rd day) * P(busy line on the 4th day)
\(= \frac{2}{3} * \frac{2}{3} * \frac{2}{3} * \frac{2}{3} = \frac{16 }{ 81}\)
Since 16/81 is less than 1/4, Quantity B is greater.
P(busy line on day 1 AND day 2 AND day 3 AND day 4) = P(busy line on the 1st day) * P(busy line on the 2nd day) * P(busy line on the 3rd day) * P(busy line on the 4th day)
\(= \frac{2}{3} * \frac{2}{3} * \frac{2}{3} * \frac{2}{3} = \frac{16 }{ 81}\)
Since 16/81 is less than 1/4, Quantity B is greater.
gmatclubot
The probability that Frances gets a line busy whenever she calls to [#permalink]
15 Nov 2024, 10:59
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