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Jonathan begins his vacation at the Lonely Traveler Beach Resort today 07 May 2017, 14:19
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E
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78% (01:18) correct 22% (02:11) wrong based on 151 sessions

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Jonathan begins his vacation at the Lonely Traveler Beach Resort today and he is staying at the resort until the end of the first day on which it rains. If there is a 40 percent chance of rain each day, what is the probability of him leaving at the end of the third day?

A. 0.6
B. 0.96
C. 0.64
D. 0.216
E. 0.144
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E
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Jonathan begins his vacation at the Lonely Traveler Beach Resort today Updated on: 19 Sep 2021, 13:14
Originally posted by BrentGMATPrepNow on 07 May 2017, 14:32.
Last edited by BrentGMATPrepNow on 19 Sep 2021, 13:14, edited 3 times in total.
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Bunuel
Jonathan begins his vacation at the Lonely Traveler Beach Resort today and he is staying at the resort until the end of the first day on which it rains. If there is a 40 percent chance of rain each day, what is the probability of him leaving at the end of the third day?

A. 0.6
B. 0.96
C. 0.64
D. 0.216
E. 0.144
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If P(rain) = 0.4, then P(NO rain) = 0.6

P(Jonathan leaves at end of Day 3) = P(no rain on Day 1 AND no rain on Day 2 AND rain on Day 3)
= P(NO rain on Day 1) x P(NO rain on Day 2) x P(rain on Day 3)
= 0.6 x 0.6 x 0.4
= 0.144

Answer: E
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Jonathan begins his vacation at the Lonely Traveler Beach Resort today 19 Jun 2017, 02:21
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Bunuel
Jonathan begins his vacation at the Lonely Traveler Beach Resort today and he is staying at the resort until the end of the first day on which it rains. If there is a 40 percent chance of rain each day, what is the probability of him leaving at the end of the third day?

A. 0.6
B. 0.96
C. 0.64
D. 0.216
E. 0.144
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3 Day Holiday
P (No Rain) = 0.6
P (Rain) = 0.4
We only want to know the case where it rains on the 3rd day, so we need to find only once case where it doesn't rain on the first 2 days, and rains on the third day.

Probability = 0.4*0.6*0.6 = 0.144

E is the answer.
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Jonathan begins his vacation at the Lonely Traveler Beach Resort today 19 Jun 2017, 11:17
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Bunuel
Jonathan begins his vacation at the Lonely Traveler Beach Resort today and he is staying at the resort until the end of the first day on which it rains. If there is a 40 percent chance of rain each day, what is the probability of him leaving at the end of the third day?

A. 0.6
B. 0.96
C. 0.64
D. 0.216
E. 0.144
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0.60*0.60*0.40 = 0.144

Answer will be (E)
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Jonathan begins his vacation at the Lonely Traveler Beach Resort today 19 Jun 2017, 15:59
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E)
Day 1 (sunshine) = 6/10
Day 2 (sunshine) = 6/10
Day 3 (rain) = 4/10

D1 * D2 * D3 = 144/1000 = 14.4%
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Jonathan begins his vacation at the Lonely Traveler Beach Resort today 20 Jun 2017, 18:49
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Jonathan begins his vacation at the Lonely Traveler Beach Resort today and he is staying at the resort until the end of the first day on which it rains. If there is a 40 percent chance of rain each day, what is the probability of him leaving at the end of the third day?

A. 0.6
B. 0.96
C. 0.64
D. 0.216
E. 0.144
Show more

We are given that there is a 40 percent chance of rain each day and thus a 60 percent chance of no rain.

We can denote N as the probability of no rain and R as the probability of rain. We need to determine:

P(N-N-R) = 0.6 x 0.6 x 0.4 = 0.144

Answer: E
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Jonathan begins his vacation at the Lonely Traveler Beach Resort today 26 Jun 2017, 03:39
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Bunuel
Jonathan begins his vacation at the Lonely Traveler Beach Resort today and he is staying at the resort until the end of the first day on which it rains. If there is a 40 percent chance of rain each day, what is the probability of him leaving at the end of the third day?

A. 0.6
B. 0.96
C. 0.64
D. 0.216
E. 0.144
Show more


It did not rain on the first two days but on the third day

rain = 0.4 ; no rain = 0.6

Desired outcome = 0.6*0.6*0.4=0.144

E
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Jonathan begins his vacation at the Lonely Traveler Beach Resort today 26 Jun 2017, 04:52
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Jonathan begins his vacation at the Lonely Traveler Beach Resort today and he is staying at the resort until the end of the first day on which it rains. If there is a 40 percent chance of rain each day, what is the probability of him leaving at the end of the third day?

P (Rain) = 0.4
P (No Rain) = 0.6

What we are looking for:

Day 1 = P (No Rain) = 0.6
Day 2 = P (No Rain) = 0.6
Day 3 = P (Rain) = 0.4

P (3rd Day) = 0.6 * 0.6 * 0.4 = 0.144

Hence, the answer is E
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Jonathan begins his vacation at the Lonely Traveler Beach Resort today 04 Jul 2019, 06:43
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