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In a certain city in September, there is a 50% chance the [#permalink]
12 Sep 2004, 18:51

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In a certain city in September, there is a 50% chance the high temperature on any day will be greater than 70 degrees. In a 5-day period, what is the probability that the high temperatures will be 69, 75, 70, 72, and 75?

I have not got any idea on this. On the exam day, probably I do something like this.
I am not too sure whether my approach is correct.
Probability = favourable events(>70)/total number of events =
3(1/2)/5(1/2) = 3/5. Because, 2/5 is closer to this, I am choosing C.

crackgmat3 wrote:

In a certain city in September, there is a 50% chance the high temperature on any day will be greater than 70 degrees. In a 5-day period, what is the probability that the high temperatures will be 69, 75, 70, 72, and 75?

In a certain city in September, there is a 50% chance the high temperature on any day will be greater than 70 degrees. In a 5-day period, what is the probability that the high temperatures will be 69, 75, 70, 72, and 75?

A. 5/13 B. 7/20 C. 2/5 D. 1/4 E. 5/16

confused a bit about this one...just because there is a 50% chance that the temp will be greater than 70 doesn't necessarily give any exact probability of having a day with a specific high temp, does it? If the question asked "In a 5 day period, what is the probability that the high temp will be <70, <70, 70, >70, >70" then would be better, but don't know when talking about specific temps...

My explanation is:
This is similar to tossing a coin five times where the probability of getting a Head or Tail is 1/2. In this case also the probability of temp greater than 70 or temp <= 70 is 1/2. Now as there are 5 trials and the two events, the total possible outcomes is 2^5 = 32
Now the given event is given as (69, 75, 70, 72, and 75) i.e the temp greater than 70 is 3 times and <=70 is 2 times.
Hence C(5,3). similar to throwing 3 heads and 2 tails in 5 trials.
Hence the required probability is
C(5,3) / 32 = 5/16

Another direct formula which can be applied is:
To find the probability of r favorable outcomes in n trials given the probabilities of occurence of the events.
Answer is: C(n,r) (p)^r * (q)^n-r
Here p+q =1
Hence applying the same here,
required probability is = 5! / (3! 2!) * (1/2)^3 * (1/2)^2 = 5/16

My explanation is: This is similar to tossing a coin five times where the probability of getting a Head or Tail is 1/2. In this case also the probability of temp greater than 70 or temp <= 70 is 1/2. Now as there are 5 trials and the two events, the total possible outcomes is 2^5 = 32 Now the given event is given as (69, 75, 70, 72, and 75) i.e the temp greater than 70 is 3 times and <=70 is 2 times. Hence C(5,3). similar to throwing 3 heads and 2 tails in 5 trials. Hence the required probability is C(5,3) / 32 = 5/16

Another direct formula which can be applied is: To find the probability of r favorable outcomes in n trials given the probabilities of occurence of the events. Answer is: C(n,r) (p)^r * (q)^n-r Here p+q =1 Hence applying the same here, required probability is = 5! / (3! 2!) * (1/2)^3 * (1/2)^2 = 5/16

I disagree with your initial thought...this question isn't phrased so that it can be similar to "flipping a coin five times"...the question doesn't ask "what is the probability that the temp will be great or less than 70 degrees", it asks about the probability that the temp will be a specific temp. So there aren't just two possible outcomes - there are significantly more - could be 52 degrees, could be 90 degrees, could be 110 degrees.

Lovely,
I think we are not concerned by what is the specific temperature, but are concerned about whether the temp is <=70 or >70. The answer would have been same if the high temperatures were 110, 200, 150, 30,70.
or for that matter provided 3 values are above 70 and other 2 are <=70

There are 2^5 possible outcomes. We are categorising those with value <=70 and >70 each having probability of 1/2.

hardworker, am I more or less correct in my understanding of the subject?

Agreed that my first explanation is a bit vague.
Still learning Probability!!

Lovely, I think we are not concerned by what is the specific temperature, but are concerned about whether the temp is <=70 or >70. The answer would have been same if the high temperatures were 110, 200, 150, 30,70. or for that matter provided 3 values are above 70 and other 2 are <=70

There are 2^5 possible outcomes. We are categorising those with value <=70 and >70 each having probability of 1/2.

hardworker, am I more or less correct in my understanding of the subject?

Agreed that my first explanation is a bit vague. Still learning Probability!!

then the problem should read differently, because just because there is a 50% chance of the temp being greater than 70 doesn't mean that there is a 50% chance of the temp being 72 or 73 or 75...

This looks like a poorly written problem since it states exact values rather than qualitative comparisons (i.e. greater than 70 degrees, less than or equal to 70 degrees).

Yes, I spent a lot of time today morning thinking on this one (how can one predict specific temperature numbers?). Agree with Hjort that this is one of those poorly worded (peterson?) problems.

I would have preferred it to be:
"In a 5-day period, what is the probability that the high temperatures are greater than 70 on exactly 3 days"

But given the conditions, I guessed that this is what they tried to test. I solved it in the first way that wireless_neo solved (similar to 2 coins flip). But the C(n,r) (p)^r * (q)^n-r formula would have been handy had the probability been different from 0.5