probability

Probability Level 3

While shuffling a pack of playing cards, four are accidently dropped.so what's the probability that are dropped are one from each suit?

52/20825 1/20825 2197/20825 1/2197

Priyam Das by Priyam Das , 27, India

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2 solutions

Michael Mendrin
Jul 14, 2014

Brian is right, it's actually 2917/20825.

But here's how to cheat on this one. Suppose an urn has equal numbers of 4 different colored marbles. I choose (and replace) 4 marbles in succession. What's the probability that I'll have chosen 4 differently colored marbles? Its

3 4 2 4 1 4 = 0.09375... \frac { 3 }{ 4 } \frac { 2 }{ 4 } \frac { 1 }{ 4 } =0.09375...

Only "2016/20825:" is even close to this, so I went with that one, even though it's incorrect.

2 Helpful 0 Interesting 0 Brilliant 0 Confused

The probability is actually ( 13 1 ) 4 ( 52 4 ) = 2197 20825 \dfrac{\binom{13}{1}^{4}}{\binom{52}{4}} = \dfrac{2197}{20825} .

2 Helpful 0 Interesting 0 Brilliant 0 Confused

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Calvin Lin Staff - 6 years, 10 months ago

Ans is incorrect as ordrr is not considered here ans should be divided by 4!

bhavya jain - 6 years, 6 months ago

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Sir, 13C1are no. Of executive ways for selecting a carf from one suit There are 4 suit Therefore fav. Outcome= (13C1)^4 Total outcomes = (52C4) Probability= (13C1)^4/52C4

Shivam K - 5 years, 6 months ago

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