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KarishmaB
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A certain bag of gemstones is composed of two-thirds 16 Mar 2025, 21:03
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Natansha
KarishmaB
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Can someone explain the

2/3 * (2R-1)/(3R-1)

part?

Say, a bag has 6 diamonds and 3 rubies. What is the probability of selecting 2 diamonds one after the other without replacement?

Probability of selecting one diamond = 6/9
Probability of selecting yet another diamond after selecting one = 5/8 (no of diamonds has gone down by 1 and total no. of diamonds has gone down by 1 too)
Total probability = (6/9)*(5/8)

Here, we assume that no of rubies is R and no of diamonds is 2R (since no of diamonds is twice the no of rubies)
Probability of selecting two diamonds without replacement = (2R/3R) * (2R - 1)/(3R - 1) = 5/12
Either cross multiply to get the value of R or try to plug in some values to see where you get a multiple of 12 in the denominator.
Once you get the value of R as 3, you know the number of diamonds is 6.

Probability of picking two rubies one after the other without replacement = (3/9) *(2/8) = 1/12
Hi Karishma, can we cross R in 2R/3R since R can't be 0 or negative?

Also since diamonds are two times as rubies, why can't the probability of ruby be 1/2 that of diamond?
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Yes, we can cancel out R from 2R/3R to get 2/3, but not from (2R - 1)/(3R - 1).
So we get 2/3 * ((2R - 1)/(3R - 1) = 5/12 which we then need to solve to get R or we plug in.

Probability of selecting 1 Ruby is 1/2 that of selecting 1 diamond. As we see here, there are 6 diamonds and 3 rubies.
Probability of selecting 1 diamond = 6/9 = 2/3
Probability of selecting 1 ruby = 3/9 = 1/3

So probability of selecting a ruby is 1/2 that of diamond.

But we are asked the probability of selecting 2 rubies which will be much lower. Think about it practically - you have to select 2 of the rare things. Its probability will be much lower than the probability of selecting 2 of the abundant things. There are those 1 abundant and 1 rare combinations possible too. Hence 1/2 doesn't work here.
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