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21 Dec 2021, 02:01
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GK002
Dear GK002
we do not care concerning the permutation.
as you alluded there 2 cases, but the calculation is different.
BG or GB = BR + GB = where
BG =\(\frac{1}{5}\)*\(\frac{4}{4}\) (because the probability to take G is 100% after you removed "B") = \(\frac{1}{5}\)
GB = \(\frac{4}{5}\)*\(\frac{1}{4}\) =\(\frac{1}{5}\)
So, BG or GB = BR + GB =\(\frac{1}{5}\)+\(\frac{1}{5}\)=\(\frac{2}{5}\)
Hope it helps
15 Jul 2022, 07:50
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Can someone help me with this doubt?
The question says 'If Henry is to select 2 apples from the basket simultaneously and at random'.
Is the treatment of the question same for 'simultaneously and without replacement?'
I thought we need to take 2 cases only when question mentions 'without replacement'.
Please help!!
The question says 'If Henry is to select 2 apples from the basket simultaneously and at random'.
Is the treatment of the question same for 'simultaneously and without replacement?'
I thought we need to take 2 cases only when question mentions 'without replacement'.
Please help!!
