I did two attempts at this problem but none of them is correct.

The problem: A weather forecaster classifies all days as wet or dry. She estimates that the probability that 1 June next year is wet is 0.4. If any particular day in June is wet, the probability that the next day is wet is 0.6; otherwise the probability that the next day is wet is 0.3. Find the probability that, next year at least one of the first three days of June is wet.

1st attempt: Probability of at least one = probability of all 3 are wet - probability none is wet = $0.4\times 0.6\times 0.6-0.6\times0.4\times0.4=0.09$

2nd attempt: Probability that 3rd day is wet given 1&2 days are wet = $\frac{0.6\times 0.4\times 0.6}{0.4\times 0.6}=0.6$
Probability that 3rd is not wet given 1&2 days are not wet = $\frac{0.3\times 0.6\times 0.3}{0.6\times 0.3}=0.3$
So, probability of at least one is wet is 0.6-0.3=0.3.

The correct solution is 0.706.

I am wondering where did I make the mistake.
Helps are greatly appreciated, many thanks!


Probability of at least one = 1 - probability of none.

Probability of none = 0.6 (first is dry) * 0.7(second is dry, given first is dry) * 0.7(third is dry, given second is dry)=0.294

  • $\begingroup$ Thanks! ah yes, should have thought about this more carefully. $\endgroup$ – user71346 Feb 20 '14 at 0:45

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