Difference between revisions of "2008 iTest Problems/Problem 36"
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==Problem== | ==Problem== | ||
| − | Let <math>c</math> be the probability that the cards are neither from the same suit or the same rank. Compute <math>\lfloor 1000c\rfloor</math>. | + | Let <math>c</math> be the probability that the cards are neither from the same suit or the same rank. Compute <math>\lfloor 1000c\rfloor</math>. |
| + | |||
| + | '''Note: Two cards are drawn.''' | ||
==Solution== | ==Solution== | ||
| − | Note that one can not get two cards with the same suit or the same rank, so the two events are independent. The probability of getting two cards with the same rank is <math>\tfrac{13 \cdot 4 \cdot 3}{52 \cdot 51} = \tfrac{3}{51}</math>, and the probability of getting two cards with the same suit is <math>\tfrac{4 \cdot 13 \cdot 12}{52 \cdot 51} = \tfrac{12}{51}</math>. Thus, the probability of getting two cards with same suit or same rank is <math>\tfrac{3}{51} + \tfrac{12}{51} = \tfrac{15}{51}</math>, so the probability of getting two cards that do not have the same suit or rank is <math>1 - \tfrac{15}{51} = \tfrac{36}{51}</math>. That means <math>\lfloor 1000c \rfloor = \boxed{ | + | Note that one can not get two cards with the same suit or the same rank, so the two events are independent. The probability of getting two cards with the same rank is <math>\tfrac{13 \cdot 4 \cdot 3}{52 \cdot 51} = \tfrac{3}{51}</math>, and the probability of getting two cards with the same suit is <math>\tfrac{4 \cdot 13 \cdot 12}{52 \cdot 51} = \tfrac{12}{51}</math>. Thus, the probability of getting two cards with same suit or same rank is <math>\tfrac{3}{51} + \tfrac{12}{51} = \tfrac{15}{51}</math>, so the probability of getting two cards that do not have the same suit or rank is <math>1 - \tfrac{15}{51} = \tfrac{36}{51}</math>. That means <math>\lfloor 1000c \rfloor = \boxed{705}</math>. |
==See Also== | ==See Also== | ||
Latest revision as of 22:50, 4 November 2020
Problem
Let 
be the probability that the cards are neither from the same suit or the same rank. Compute 
.
Note: Two cards are drawn.
Solution
Note that one can not get two cards with the same suit or the same rank, so the two events are independent. The probability of getting two cards with the same rank is 
, and the probability of getting two cards with the same suit is 
. Thus, the probability of getting two cards with same suit or same rank is 
, so the probability of getting two cards that do not have the same suit or rank is 
. That means 
.
See Also
| 2008 iTest (Problems) | ||
| Preceded by: Problem 35 |
Followed by: Problem 37 | |
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