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A certain tin contains exactly 10 identically shaped gumdrop
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Carcass
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Joined: 18 Apr 2015
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GRE 1: Q160 V160
A certain tin contains exactly 10 identically shaped gumdrop [#permalink] New post  11 Aug 2018, 09:44
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A certain tin contains exactly 10 identically shaped gumdrops, of which 2 are cherry, 2 are lemon, 2 are grape, 2 are cinnamon and 2 are strawberry. If 2 gumdrops are randomly selected, one at a time and without replacement, what is the probability that the second gumdrop is the same flavor as the first?

A. \(\frac{1}{45}\)

B. \(\frac{1}{25}\)

C. \(\frac{1}{9}\)

D. \(\frac{1}{5}\)

E. \(\frac{2}{5}\)
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Re: A certain tin contains exactly 10 identically shaped gumdrop [#permalink] New post  11 Aug 2018, 10:03
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Carcass wrote:
A certain tin contains exactly 10 identically shaped gumdrops, of which 2 are cherry, 2 are lemon, 2 are grape, 2 are cinnamon and 2 are strawberry. If 2 gumdrops are randomly selected, one at a time and without replacement, what is the probability that the second gumdrop is the same flavor as the first?

A. \(\frac{1}{45}\)

B. \(\frac{1}{25}\)

C. \(\frac{1}{9}\)

D. \(\frac{1}{5}\)

E. \(\frac{2}{5}\)





Here,
for the first draw we pick any gumdrop,

Now we have remaining 9 gumdrop and selection a matching flavor with the first one = \(\frac{1}{9}\)
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kunalkmr62
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Re: A certain tin contains exactly 10 identically shaped gumdrop [#permalink] New post  24 Nov 2018, 06:01
We also need to calculate the probability of selecting the 1st gumdrop , which will be 1/5 for any gumdeop and then multiply the probability of selecting the 2nd gumdrop , which should be 1/9.
To make it clear would you have also ignored the probability of selecting the 1st gumdrop if the numbers would have been different ?
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chetan2u
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Re: A certain tin contains exactly 10 identically shaped gumdrop [#permalink] New post  24 Nov 2018, 06:31
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kunalkmr62 wrote:
We also need to calculate the probability of selecting the 1st gumdrop , which will be 1/5 for any gumdeop and then multiply the probability of selecting the 2nd gumdrop , which should be 1/9.
To make it clear would you have also ignored the probability of selecting the 1st gumdrop if the numbers would have been different ?



No...

You can choose any of the first so 10/10, the second has to be a pair of this so1/9
Total 10/10*(1/9)=1/9..

If we take the way you have taken..
Then you are calculating ways to pick up a specific gumdeop..
So (2/10)*(1/9)...
But the one you choose can be any of five so 5C1..
So 5C1*(1/5)(1/9)=5*(1/5)(1/9)=1/9
Same answer
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