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#1
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Solving problems in Reverse
If I want to calculate the probability of having at least one diamond in a draw of two cards -- I will subtract from 1 the probability of having no diamonds in two cards,
1 - (39/52 * 38/51) = 1 - .5588 = .4412 What if I wanted to work the problem forward instead of backwards -- what would be the easiest way? |
#2
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Re: Solving problems in Reverse
The answer depends on whether you would accept an approximation and how accurate you want the approximation to be e.g. my answer to all problems is 1 [img]/images/graemlins/wink.gif[/img]
The most obvious "forward" way as you call it would be: 13/52 + (39/52)*(13/51) It is always possible to factor out the 13/52 in this type of problem i.e. <font class="small">Code:</font><hr /><pre>x/y + ((y - x)/y)*(x/(y - 1)) = x/y + (y - x)*x/((y - 1)*y) = x/y + ((y - x)/(y - 1))*(x/y) = (x/y)*(1 + (y - x)/(y - 1))</pre><hr /> This would look like: (13/52)*(1 + 39/51) in the problem at hand. It is about as economical to calculate as the "reverse" way, but has drifted away from how most people think. |
#3
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Re: Solving problems in Reverse
A is the event that the first card is a diamond but the second card isn't .
B is the event that the second card is a diamond but the first isn't . C is the event that both cards are diamonds . We simply add these 3 mutually exclusive events . P(A)= 13/52* 39/51 = P(B) P(C) = 13/52*12/51 P(A) + P(B) + P(C) = 0.4412 |
#4
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Re: Solving problems in Reverse
Here is a probabilistic explanation for R. Gilbert's answer .
13/52 is the probability the first card is a diamond . This may also include the probability when the second card is a diamond or not . The probability the first card isn't a diamond but the second card is : 39/52*13/51 . These two events are mutually exclusive and we can add them directly . |
#5
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Re: Solving problems in Reverse
Thank you both for your answers. I know I asked for the easiest way (my bad) -- reviewing the two answers I should have said the most intuitive way. Thank you R Gilbert for the easiest -- Thank you jay_shark for the most intuitive and an explanation of the easiest.
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