My Math Forum Probability(hard question)

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 April 19th, 2011, 11:51 PM #1 Senior Member   Joined: Dec 2010 Posts: 233 Thanks: 0 Probability(hard question) Today my teacher gave us a topical test on probability.this is one of the question. In a box containing 20 red apples and 5 green apples,3 of the red apples are rotten and one of the green apples is rotten.Two apples are randomly selected from the box.Calculate the probability that a)both the apples are rotten b)both are red and at least one of them is rotten c)at least one is rotten if both are red d)both are red if it is found that at least one is rotten I construct a tree diagram for this question.i think i am wrong,right?i am confuse about this question.Please help me to solve step by step and explain to me about the concept.Thank you.
April 20th, 2011, 05:04 AM   #2
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Re: Probability(hard question)

Hello, hoyy1kolko!

Quote:
 In a box containing 20 red apples and 5 green apples, 3 of the red apples are rotten and one of the green apples is rotten. Two apples are randomly selected from the box.

$\text{There are: }\:{25\choose2} \,=\,300\text{ possible outcomes.}$

Quote:
 Calculate the probability that: a) both the apples are rotten

$\text{W\!e want 2 of the 4 rotten apples: }\:{4\choose2} \,=\,6\text{ ways.}$

$\text{Therefore: }\:P(\text{both rotten}) \:=\:\frac{6}{300} \:=\:\frac{1}{50}$

Quote:
 b) both are red and at least one of them is rotten

$\text{Both red: }\:{20\choose2} \,=\,190\text{ ways}$

$\text{No rotten red apples: }\:{17\choose2} \,=\,136\text{ ways}$

$\text{Hence: \,Both red and at least one rotten} \:=\:190\,-\,136 \:=\:54\text{ ways.}$

$\text{Therefore: }\:P(\text{both red }\wedge\text{ at least one rotten}) \:=\:\frac{54}{300} \:=\:\frac{9}{50}$

Quote:
 c) at least one is rotten if both are red

$\text{W\!e want: }\:P(\text{at least 1 rotten }|\text{ both red})$

$\text{Bayes' Theorem: }\;P(\text{at least 1 rotten }|\text{ both red}) \;=\;\frac{P(\text{at least 1 rotten }\wedge\text{ both red})}{P(\text{both red})}$

$\text{W\!e found the numerator in part (b): }\:\frac{9}{50}$

$\text{The denominator is: }\:P(\text{both red}) \:=\:\frac{{20\choose2}}{300} \:=\:\frac{190}{300} \:=\:\frac{19}{30}$

$\text{Therefore: }\:P(\text{at least 1 rotten }|\text{ both red}) \:=\:\frac{\frac{9}{50}}{\frac{19}{30}} \:=\:\frac{27}{95}$

Quote:
 d) both are red if it is found that at least one is rotten

$\text{W\!e want: }\:P(\text{both red }|\text{ at least 1 rotten}) \;=\;\frac{P(\text{both red }\wedge\text{ at least 1 rotten})}{P(\text{at least 1 rotten})}$

$\text{The numerator is the same as (c): }\,\frac{9}{50}$

$\text{The opposite of "at least 1 rotten" is "}no\text{ rotten".}$
[color=beige]. . [/color]$\text{W\!e want 2 of the 21 good apples: }\:{21\choose2} \,=\,210\text{ ways.}$
[color=beige]. . [/color]$\text{Then: }\:P(\text{no rotten}) \:=\:\frac{210}{300} \:=\:\frac{7}{10}$
$\text{Hence: }\:P(\text{at least 1 rotten}) \:=\:1\,-\,\frac{7}{10}\:=\:\frac{3}{10}$

$\text{Therefore: }\:P(\text{both red }|\text{ at least 1 rotten}) \;=\;\frac{\frac{9}{50}}{\frac{3}{10}} \;=\;\frac{3}{5}$

 April 21st, 2011, 07:04 AM #3 Senior Member   Joined: Dec 2010 Posts: 233 Thanks: 0 Re: Probability(hard question) Thank you so much sir,yr explanation is excellent and easy to understand.

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