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Consider the experiment of drawing two cards from a deck in which all picture cards have been removed and adding their values (with ace = $1$).

  • What is the sample space?

  • What is the probability of obtaining a total of $5$ for the two cards?

  • Let A be the event “total card value is 5 or less.” Find $P ( A )$

  • Let A be the event “total card value is 5 or less.” Find $P (A^{\complement} )$

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  • $\begingroup$ and adding their values (with ace = 1) means? are only face cards are removed or Ace as well? $\endgroup$ – idea Nov 16 '18 at 17:39
  • $\begingroup$ The question in the header doesn't appear to match the question(s) in the body. $\endgroup$ – lulu Nov 16 '18 at 17:44
  • $\begingroup$ Welcome to Math.SE. Take a look at both links How to ask a good question at Math.SE and for formatting MathJax. To avoid downvotes and closing you should add your own efforts to the question, and tell us where you got stuck. $\endgroup$ – drhab Nov 16 '18 at 17:58
  • $\begingroup$ header should have said 5 and not 6 and probably not the best title. I apologize for confusion. The Ace = 1 and jack, queen, and king should be 11, 12, 13 respectfully. $\endgroup$ – user1222339 Nov 16 '18 at 18:12
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Assuming that all face cards are removed, you are left with $52-12=40$ cards.

So, your Sample Space is $40$ cards consisting of all numbered cards (from Ace=$1$ to $10$) in all $4$ suits.
When you draw $2$ cards to obtain a total of $5$, there are following possibilities: $$(4,1),(3,2)$$ and there are $4$ cards of each kind.
So select $1$ of them in $^4C_1$ way, and $2$ cards can be selected from $40$ in $^{40}C_2$ ways. $$P(\text{sum=5})=\frac{^4C_1\cdot ^4C_1+^4C_1\cdot ^4C_1}{^{40}C_2}=\frac{32}{^{40}C_2}$$ For $sum\leq5$, possible cases are: $$(1,1),(1,2),(1,3),(2,2),(1,4),(2,3)$$ There are $^4C_2$ ways to draw $2$ cards with same number, from $4$. $$P(A)=\frac{(^4C_2)+(^4C_1\cdot ^4C_1)+(^4C_1\cdot ^4C_1)+(^4C_2)+(^4C_1\cdot ^4C_1)+(^4C_1\cdot ^4C_1)}{^{40}C_2}=\frac{76}{^{40}C_2}$$ $$P(A^C)=1-P(A)=1-\frac{76}{^{40}C_2}$$

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  • $\begingroup$ The answer seems correct to me, but can you help me understand what the C stands for? what is the value? Also shouldn't it be 5 and not 4 to the left of the C? since it is the sum less than or equal to 5? $\endgroup$ – user1222339 Nov 16 '18 at 19:49
  • $\begingroup$ C stands for Combinations. $^nC_r$ means number of ways to select r objects from n. $$^nC_r=\frac{n!}{r! \cdot (n-r)!}$$ There, 4 represents the number of available cards having same number, eg. You have 4 1's out of which you select 1 or more. Just take a lecture on Combinations and you will understand the solution fully. $\endgroup$ – idea Nov 17 '18 at 6:30

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