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The probability of rain on each of five days is 1/6, except on the fir

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The probability of rain on each of five days is 1/6, except on the fir  [#permalink]

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New post 18 Jul 2017, 21:19
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A
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C
D
E

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  35% (medium)

Question Stats:

78% (02:17) correct 22% (02:28) wrong based on 136 sessions

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Re: The probability of rain on each of five days is 1/6, except on the fir  [#permalink]

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New post 18 Jul 2017, 21:25
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Bunuel wrote:
The probability of rain on each of five days is 1/6, except on the first day, when it is 2/5, and on the last day, when it is 4/5. What is the probability that rain will occur on at least one of the five days?

(A) 1/675
(B) 5/72
(C) 5/27
(D) 22/27
(E) 67/72


No rain on all days
\(3/5*5/6*5/6*5/6*1/5 = 15/216\)

Atleast 1 day rain = \(1-15/216 = 201/216 = 67/72\)
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The probability of rain on each of five days is 1/6, except on the fir  [#permalink]

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New post 18 Jul 2017, 21:29
1
Given data :
Probability of raining on 1st day = \(\frac{2}{5}\)
Probability of raining on other days = \(\frac{1}{6}\)
Probability of raining on 5th day = \(\frac{4}{5}\)

From this we can find probability it didn't rain on
Day 1 = 1 - \(\frac{2}{5}\) = \(\frac{3}{5}\)
Day 2,3,4 = 1 - \(\frac{1}{6}\) = \(\frac{5}{6}\)
Day 5 = 1 - \(\frac{4}{5}\) = \(\frac{1}{5}\)

Since we have been asked to find the probability of rain on at least 1 days,
the easiest was is to find the probability that it doesn't rain on either of the days.

P(Raining on at least one day) = 1 - P(Not raining on single day)
= 1 - (\(\frac{3}{5}*\frac{5}{6}*\frac{5}{6}*\frac{5}{6}*\frac{1}{5}\)) = 1 - \(\frac{5}{72} = \frac{67}{72}\)(Option E)
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Re: The probability of rain on each of five days is 1/6, except on the fir  [#permalink]

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New post 18 Jul 2017, 21:55
Bunuel wrote:
The probability of rain on each of five days is 1/6, except on the first day, when it is 2/5, and on the last day, when it is 4/5. What is the probability that rain will occur on at least one of the five days?

(A) 1/675
(B) 5/72
(C) 5/27
(D) 22/27
(E) 67/72


Probability of rain on at least one of the days = 1 - (Probability of rain on no day).

Probability of rain on Day-1 = \(\frac{2}{5}\), no rain = \(\frac{3}{5}\).
Probability of rain on Day-2 = Day-3 = Day-4 = \(\frac{1}{6}\), no rain = \(\frac{5}{6}\).
Probability of rain on Day-4 = \(\frac{4}{5}\), no rain = \(\frac{1}{5}\).

Probability of rain on at least one of the days = \(1 - (\frac{3}{5} * (\frac{5}{6})^{3} * \frac{1}{5}) = 1 - \frac{15}{216} = 1 - \frac{5}{72} = \frac{67}{72}\). Ans - E.
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Re: The probability of rain on each of five days is 1/6, except on the fir  [#permalink]

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New post 18 Jul 2017, 22:31
Probability that rain will not happen on any day is-

=(1-2/5)(1-1/6)(1-1/6)(1-1/6)(1-4/5)
=3/5*5/6*5/6*5/6*1/5
=5/72

now for probability of rain to occur on at least one of five days, We will subtract probability of not occurring into 1 i.e

=1-5/72
=67/72

Answer is E
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Re: The probability of rain on each of five days is 1/6, except on the fir  [#permalink]

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New post 13 Oct 2017, 23:08
What if it was asked to find the probability of raining for atleast 2 days??

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Re: The probability of rain on each of five days is 1/6, except  [#permalink]

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Re: The probability of rain on each of five days is 1/6, except   [#permalink] 29 Apr 2019, 22:55
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