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Probability - By Counting

For COMPETITION
Number of Total Problems: 7.
FOR PRINT ::: (Book)

Problem Num : 1

From : NCTM

Type: None

Section:Probability

Theme:None
Adjustment# :

Difficulty: 1

Category by Counting

Analysis

Solution/Answer

Problem Num : 2

From : NCTM

Type: Understanding

Section:Probability

Theme:None
Adjustment# :

Difficulty: 2

Category by Counting

Analysis

Solution/Answer

Problem Num : 3

From : NCTM

Type: None

Section:Probability

Theme:None
Adjustment# :

Difficulty: 1

Category by Counting

Analysis

Solution/Answer

Problem Num : 4

From : NCTM

Type: Understanding

Section:Probability

Theme:None
Adjustment# :

Difficulty: 2

Category by Counting

Analysis

Solution/Answer

Problem Num : 5

From : NCTM

Type: None

Section:Probability

Theme:None
Adjustment# :

Difficulty: 1

Category by Counting

Analysis

Solution/Answer

Problem Num : 6

From : AMC10B

Type:

Section:Probability

Theme:
Adjustment# : 0

Difficulty: 1

'
Forty slips are placed into a hat, each bearing a number $1$ , $2$ , $3$ , $4$ , $5$ , $6$ , $7$ , $8$ , $9$ , or $10$ , with each number entered on four slips. Four slips are drawn from the hat at random and without replacement. Let $p$ be the probability that all four slips bear the same number. Let $q$ be the probability that two of the slips bear a number $a$ and the other two bear a number $b eq a$ . What is the value of $q/p$ ?
$mathrm{(A)} 162 qquad mathrm{(B)} 180 qquad mathrm{(C)} 324 qquad mathrm{(D)} 360 qquad mathrm{(E)} 720$
'

Category By Counting

Analysis

Solution/Answer
There are $10$ ways to determine which number to pick. There are $4!$ way to then draw those four slips with that number, and $40 cdot 39 cdot 38 cdot 37$ total ways to draw four slips. Thus $p = frac{10cdot 4!}{40 cdot 39 cdot 38 cdot 37}$ .
There are ${10 choose 2} = 45$ ways to determine which two numbers to pick for the second probability. There are ${4 choose 2}$ ways to arrange the order which we draw the non-equal slips, and in each order there are $4 imes 3 imes 4 imes 3$ ways to pick the slips, so $q = frac{45 cdot 6 cdot 4^2 cdot 3^2}{40 imes 39 imes 38 imes 37}$ .
Hence, the answer is $frac{q}{p} = frac{2^5 cdot 3^4 cdot 5}{10cdot 4!} = 162 mathbf{(A)}$ .

Answer:

Problem Num : 7

From : AMC10

Type:

Section:Probability

Theme:
Adjustment# : 0

Difficulty: 1

'
A set of three points is randomly chosen from the grid shown. Each three point set has the same probability of being chosen. What is the probability that the points lie on the same straight line?

$mathrm{(A) } frac{1}{21} qquad mathrm{(B) } frac{1}{14} qquad mathrm{(C) } frac{2}{21} qquad mathrm{(D) } ...$
'

Category By Counting

Analysis

Solution/Answer
There are $inom{9}{3}$ ways to choose three points out of the 9 there. There are 8 combinations of dots such that they lie in a straight line: three vertical, three horizontal, and the diagonals.
$dfrac{8}{inom{9}{3}}=dfrac{8}{84}=dfrac{2}{21} Rightarrow (C)$

Answer:

Array ( [0] => 3690 [1] => 3713 [2] => 3717 [3] => 3757 [4] => 3772 [5] => 8044 [6] => 7791 ) 7