SOLUTION: Suppose that two cards are selected at random from a standard 52 card deck. What is the probability that both cards are less than 10 and neither of them is red?
Total no of o
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Question 182417This question is from textbook discrete mathematical structures
: Suppose that two cards are selected at random from a standard 52 card deck. What is the probability that both cards are less than 10 and neither of them is red?
Total no of outcomes = 52 C 2 = 1326
there are 26 black and 26 red cards .
After that i don't know how to proceed please help me.
Thanks
This question is from textbook discrete mathematical structures
Found 3 solutions by stanbon, scott8148, solver91311:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Suppose that two cards are selected at random from a standard 52 card deck. What is the probability that both cards are less than 10 and neither of them is red?
Total no of outcomes = 52 C 2 = 1326
there are 26 black and 26 red cards .
After that i don't know how to proceed please help me.
----------------------------------------------------------
P(two<10 AND black) = P(two<10|black)*P(black)
# of black is 26
# of those black that are less then 10 = 9
---
So, P(two<10 | black) = 9C2/26C2
And P(black) = 1/2
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Ans: [9C2/26C2] = [36/325] = 18/325
=====================================================
Cheers,
Stan H.
Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!
how many cards are less than 10?
32
how many of those are not red?
16
so 16C2 is the number of possible desired outcomes
probability is (16C2)/(52C2)
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Stan is correct as long as you consider the Ace to be smaller than a 10. Your problem didn't say, so it could be either way. If Ace is larger than 10, modify Stan's answer to be:
\left(\frac{1}{2}\right) = \left(\frac{28}{325}\right)(\frac{1}{2}\right) = \frac{14}{325} )
John
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