Uniform distribution vs conditional probability

Muffins

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Hey guys! Excited to be here again as I always get very helpful responses which clarify the subjects alot to me!

This time I have a question in regards to this-

Ron is moving. He packed his clothes in 5 boxes. Upon unpacking, Ron began searching for his shirt. He chooses a random box, checks it, and if he doesn't find it there, he moves on to another box (randomly picked), until the shirt is found. What's the probability that he will find his shirt in the fourth box he'd checked?

My immediate thought was that the probability is 1/5*1/4*1/3*1/2. But looking at the answer sheet it said- if we define x to be the boxes he's checked, x~U(1,5) and thus when x=4 the probability is 1/5.

Basically I wonder if this is just a matter of how the question was phrased, because it seemed to me as if he'd carry on to check 'other' boxes, as in the remaining ones.

Would appreciate any thought about this.

Thank you kindly,
 
Hey guys! Excited to be here again as I always get very helpful responses which clarify the subjects alot to me!

This time I have a question in regards to this-

Ron is moving. He packed his clothes in 5 boxes. Upon unpacking, Ron began searching for his shirt. He chooses a random box, checks it, and if he doesn't find it there, he moves on to another box (randomly picked), until the shirt is found. What's the probability that he will find his shirt in the fourth box he'd checked?

My immediate thought was that the probability is 1/5*1/4*1/3*1/2. But looking at the answer sheet it said- if we define x to be the boxes he's checked, x~U(1,5) and thus when x=4 the probability is 1/5.

Basically I wonder if this is just a matter of how the question was phrased, because it seemed to me as if he'd carry on to check 'other' boxes, as in the remaining ones.

Would appreciate any thought about this.

Thank you kindly,
There are two ways to get the answer. The "easy" way is to say it is equally probable that the shirt is in any box number x, so P(x=4) = 1/5

If you want to go through the "chain," you have to look also at the "not" probabilities. If he DID find it on the fourth try, he must have NOT found it three times:
P(n=4) = (4/5)(3/4)(2/3) * (1/2) = 1/5

The easy way is easier - but I was glad that the product of probabilities gave the same result.
 
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