# What is the probability of obtaining a total of 6 for the two cards?

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 $$P (A^{\complement} )$$

• and adding their values (with ace = 1) means? are only face cards are removed or Ace as well? – idea Nov 16 '18 at 17:39
• The question in the header doesn't appear to match the question(s) in the body. – lulu Nov 16 '18 at 17:44
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• 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. – user1222339 Nov 16 '18 at 18:12

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 $$4$$. $$P(A)=\frac{(^4C_2)+(^4C_1\cdot ^4C_1)+(^4C_1\cdot ^4C_1)+(^4C_2)+(^4C_1\cdot ^4C_1)+(^4C_1\cdot ^4C_1)}{^{40}C_2}=\frac{76}{^{40}C_2}$$ $$P(A^C)=1-P(A)=1-\frac{76}{^{40}C_2}$$
• 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. – idea Nov 17 '18 at 6:30