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The probability of rainfall in City X on any given day is 30%. The pro
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24 Jan 2023, 09:49
24 Jan 2023, 09:49
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Question Stats:
50% (01:19) correct
50% (01:13) wrong based on 32 sessions
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The probability of rainfall in City X on any given day is 30%. The probability of rainfall on any given day is independent of whether it rains on any other day.
A. The quantity in Column A is greater.
B. The quantity in Column B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the information given
Quantity A |
Quantity B |
The probability of rainfall in City X on at least one day out of two days |
The probability of no rainfall in City X on either of those two days |
A. The quantity in Column A is greater.
B. The quantity in Column B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the information given
Re: The probability of rainfall in City X on any given day is 30%. The pro
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25 Jan 2023, 02:38
25 Jan 2023, 02:38
1
Expert Reply
OE
The probability of rainfall on at least one of the two days is equal to P(Rain 1st day) + P(Rain 2nd day) - P(Rain both days), since rainfall on each day is independent of what happened on other days. This probability equals: 0.3 + 0.3 - (0.3 x0.3) = 0.6 - 0.09 = O.51 or 51%.
Alternatively, you can use the “1 - x” trick here: rainfall on at least one of the two days is 1 - the probability of rain on either day (0.7^2). Therefore, the probability is 1-0 7 x 0 .7 = 1- 0.49 = 0.31 = 51%.
Quantity B, the probability of no rainfall on either day, follows from the work above: 0.7 x 0.7 = 0.49, which is 49%.Thus. Quantity A is greater.
The probability of rainfall on at least one of the two days is equal to P(Rain 1st day) + P(Rain 2nd day) - P(Rain both days), since rainfall on each day is independent of what happened on other days. This probability equals: 0.3 + 0.3 - (0.3 x0.3) = 0.6 - 0.09 = O.51 or 51%.
Alternatively, you can use the “1 - x” trick here: rainfall on at least one of the two days is 1 - the probability of rain on either day (0.7^2). Therefore, the probability is 1-0 7 x 0 .7 = 1- 0.49 = 0.31 = 51%.
Quantity B, the probability of no rainfall on either day, follows from the work above: 0.7 x 0.7 = 0.49, which is 49%.Thus. Quantity A is greater.
Re: The probability of rainfall in City X on any given day is 30%. The pro
[#permalink]
16 Dec 2023, 12:58
16 Dec 2023, 12:58
2
As it's a problem of independent probabilities, we may classify the problem as follows.
Let's think about the two days as Monday and Tuesday. so the independent probabilities will be--
1) Rain on Monday* No rain on Tuesday=(.3*.7)=.21
2) Rain on Tuesday * No rain on Monday=(.3*.7)=.21
3) Rain on Both day=(.3*.3)=.09
4) Neither of the day=(.7*.7)=.49
1,2,and 3(.21+.21+.09=.51) will be considered for 'at least' issue and # 4(=.49) is for neither issue. .51>.49, so the answer is A.
Let's think about the two days as Monday and Tuesday. so the independent probabilities will be--
1) Rain on Monday* No rain on Tuesday=(.3*.7)=.21
2) Rain on Tuesday * No rain on Monday=(.3*.7)=.21
3) Rain on Both day=(.3*.3)=.09
4) Neither of the day=(.7*.7)=.49
1,2,and 3(.21+.21+.09=.51) will be considered for 'at least' issue and # 4(=.49) is for neither issue. .51>.49, so the answer is A.
