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1) Out of 90 tins, 24 contained corns, 12 contained beans, 36 contained peas and the rest conatined mushrooms. Find the probability that Sean will have to open more than 2 tins before he finds one which does not contain baby corns.

 

I am not very sure how to tackle this question. I thought of using complement, but I don't know what is the eact opposite of this situation. Actually I don't really understand what the question means.

The answer was 0.069.

 

2) A box contains 6 black balls, 4 red balls and 2 yellow balls. 3 are drawn with replacement. Calculate the probability that all 3 balls are of the same colour.

 

Why isn't it (0.5)^3 x (1/3)^3?

The correct answer was 1/6.

 

3) What is the probability of getting all three cards being red out of a ordinary pack of 52 playing cards?

 

Why isn't it (26/52)x(25/51)x(24/50)?

The correct answer was 0.125.

 

Please help me understand where I am wrong! I have been constantly making mistakes like that :doh:, but I don't know what is my error in thinking. Thank you!

Posted
1) Out of 90 tins, 24 contained corns, 12 contained beans, 36 contained peas and the rest conatined mushrooms. Find the probability that Sean will have to open more than 2 tins before he finds one which does not contain baby corns.

 

I am not very sure how to tackle this question. I thought of using complement, but I don't know what is the eact opposite of this situation. Actually I don't really understand what the question means.

The answer was 0.069.

 

2) A box contains 6 black balls, 4 red balls and 2 yellow balls. 3 are drawn with replacement. Calculate the probability that all 3 balls are of the same colour.

 

Why isn't it (0.5)^3 x (1/3)^3?

The correct answer was 1/6.

 

3) What is the probability of getting all three cards being red out of a ordinary pack of 52 playing cards?

 

Why isn't it (26/52)x(25/51)x(24/50)?

The correct answer was 0.125.

 

Please help me understand where I am wrong! I have been constantly making mistakes like that :doh:, but I don't know what is my error in thinking. Thank you!

 

I think I can answer these bast in reverse order:

 

your answer the #3 is correct in most typical situations, that is, being dealt 3 cards in a row and looking at all three in your hand and the chances of them all being red. But, the answer obviously implies that it was sampling with replacement, because the chance of a red card out of a normal 52 card deck is one half and the answer given is (1/2)^3. So, there is probably a mistake in the question statement of #3

 

#2 can probably be easiest seen if you break the question statement down into its component parts:

 

Turn "what is the probability of drawing the same color three times in a row" into "what is the prob. of drawing black 3 times in a row + what is the prob. of drawing red 3 times in a row + what is the prob. of drawing yellow 3 times in a row"

 

Calculate each of the three in the right hand side, and then add them up and see if that doesn't get your answer.

 

#1 is actually going to be just like your answer (not the book's answer) to #3. Do the same thing -- find what the chances are of not opening corn the first time, then what the chances are of not opening corn the second time (which changes because it is sampling without replacement).

 

I hope this helps.

Posted

1. its best to do a tree diagram, find the p of 2 then find the p of opening 3 and thats one way. Other way is finding it indirectly like bignose said

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