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Probability - Multiplication and Addition rule

For COMPETITION
Number of Total Problems: 14.
FOR PRINT ::: (Book)

Problem Num : 1

From : NCTM

Type: None

Section:Probability

Theme:None
Adjustment# :

Difficulty: 1

Category Multiplication and Addition rule

Analysis

Solution/Answer

Problem Num : 2

From : NCTM

Type: None

Section:Probability

Theme:None
Adjustment# :

Difficulty: 1

Category Multiplication and Addition rule

Analysis

Solution/Answer

Problem Num : 3

From : NCTM

Type: Complex

Section:Probability

Theme:Case Devision
Adjustment# :

Difficulty: 2

Category Multiplication and Addition rule

Analysis

Solution/Answer

Problem Num : 4

From : NCTM

Type: Complex

Section:Probability

Theme:Case Devision
Adjustment# :

Difficulty: 2

Category Multiplication and Addition rule

Analysis

Solution/Answer

Problem Num : 5

From : NCTM

Type: Understanding

Section:Probability

Theme:Case Devision
Adjustment# :

Difficulty: 2

Category Multiplication and Addition rule

Analysis

Solution/Answer

Problem Num : 6

From : AMC10B

Type:

Section:Probability

Theme:
Adjustment# : 0

Difficulty: 1

'
A bag contains two red beads and two green beads. You reach into the bag and pull out a bead, replacing it with a red bead regardless of the color you pulled out. What is the probability that all beads in the bag are red after three such replacements?
$extbf{(A) } frac{1}{8} qquad extbf{(B) } frac{5}{32} qquad extbf{(C) } frac{9}{32} qquad extbf{(D) } frac{3}{8} ...$
'

Category Multiplication and Addition rule

Analysis

Solution/Answer
We can divide the case of all beads in the bag being red after three replacements into three cases.
The first case is in which the two green beads are the first two beads to be chosen. The probability for this is $frac{2}{4} imes frac{1}{4} imes frac{4}{4} = frac{1}{8}$
The second case is in which the green beads are chosen first and third. The probability for this is $frac{2}{4} imes frac{3}{4} imes frac{1}{4} = frac{3}{32}$
The third case is in which the green beads are chosen second and third. The probability for this is $frac{2}{4} imes frac{2}{4} imes frac{1}{4} = frac{1}{16}$
Add all these cases together $frac{1}{8}+frac{3}{32}+frac{1}{16} = frac{4}{32}+frac{3}{32}+frac{2}{32} = oxed{ extbf{(C) } frac{9}{32}}$

Answer:

Problem Num : 7

From : AMC10B

Type:

Section:Probability

Theme:
Adjustment# : 0

Difficulty: 1

'
One fair die has faces $1$ , $1$ , $2$ , $2$ , $3$ , $3$ and another has faces $4$ , $4$ , $5$ , $5$ , $6$ , $6$ . The dice are rolled and the numbers on the top faces are added. What is the probability that the sum will be odd?
$mathrm{(A)} frac{1}{3} qquad mathrm{(B)} frac{4}{9} qquad mathrm{(C)} frac{1}{2} qquad mathrm{(D)} frac{5}{9} qqu...$
'

Category Multiplication and Addition rule

Analysis

Solution/Answer
In order to obtain an odd sum, exactly one out of the two dice must have an odd number. We can easily find the total probability using casework.
Case 1: The first die is odd and the second die is even.
The probability of this happening is $dfrac{4}{6} imesdfrac{4}{6}=dfrac{16}{36}=dfrac{4}{9}$
Case 2: The first die is even and the second die is odd.
The probability of this happening is $dfrac{2}{6} imesdfrac{2}{6}=dfrac{4}{36}=dfrac{1}{9}$
Adding these two probabilities will give us our final answer. $dfrac{4}{9}+dfrac{1}{9}=oxed{mathrm{(D)} dfrac{5}{9}}$

Answer:

Problem Num : 8

From : AMC10B

Type:

Section:Probability

Theme:
Adjustment# : 0

Difficulty: 1

'
Twelve fair dice are rolled. What is the probability that the product of the numbers on the top faces is prime?
$mathrm{(A)} left(frac{1}{12} ight)^{12} qquad mathrm{(B)} left(frac{1}{6} ight)^{12} qquad mathrm{(C)} 2left(fra...$
'

Category Multiplication and Addition rule

Analysis

Solution/Answer
In order for the product of the numbers to be prime, $11$ of the dice have to be a $1$ , and the other die has to be a prime number. There are $3$ prime numbers ( $2$ , $3$ , and $5$ ), and there is only one $1$ , and there are $dbinom{12}{1}$ ways to choose which die will have the prime number, so the probability is $dfrac{3}{6} imesleft(dfrac{1}{6} ight)^{11} imesdbinom{12}{1} = dfrac{1}{2} imesleft(dfrac{1}{6} ight)^{11} imes...$ .

Answer:

Problem Num : 9

From : AMC10B

Type:

Section:Probability

Theme:
Adjustment# : 0

Difficulty: 1

'
An envelope contains eight bills: $2$ ones, $2$ fives, $2$ tens, and $2$ twenties. Two bills are drawn at random without replacement. What is the probability that their sum is $ $20$ or more?
$mathrm{(A)} {{{frac{1}{4}}}} qquad mathrm{(B)} {{{frac{2}{7}}}} qquad mathrm{(C)} {{{frac{3}{7}}}} qquad mathrm{...$
'

Category Multiplication and Addition rule

Analysis

Solution/Answer
The only way to get a total of $ $20$ or more is if you pick a twenty and another bill, or if you pick both tens. There are a total of $dbinom{8}{2}=dfrac{8 imes7}{2 imes1}=28$ ways to choose $2$ bills out of $8$ . There are $12$ ways to choose a twenty and some other non-twenty bill. There is $1$ way to choose both twenties, and also $1$ way to choose both tens. Adding these up, we find that there are a total of $14$ ways to attain a sum of $20$ or greater, so there is a total probability of $dfrac{14}{28}=oxed{mathrm{(D)} dfrac{1}{2}}$ .

Answer:

Problem Num : 10

From : AMC10

Type:

Section:Probability

Theme:
Adjustment# : 0

Difficulty: 1

'
Coin $A$ is flipped three times and coin $B$ is flipped four times. What is the probability that the number of heads obtained from flipping the two fair coins is the same?
$mathrm{(A) } frac{29}{128} qquad mathrm{(B) } frac{23}{128} qquad mathrm{(C) } frac14 qquad mathrm{(D) } f...$
'

Category Multiplication and Addition rule

Analysis

Solution/Answer
There are 4 ways that the same number of heads will be obtained; 0, 1, 2, or 3 heads.
The probability of both getting 0 heads is $left(frac12 ight)^3{3choose0}left(frac12 ight)^4{4choose0}=frac1{128}$ .
The probability of both getting 1 head is $left(frac12 ight)^3{3choose1}left(frac12 ight)^4{4choose1}=frac{12}{128}$
The probability of both getting 2 heads is $left(frac12 ight)^3{3choose2}left(frac12 ight)^4{4choose2}=frac{18}{128}$
The probability of both getting 3 heads is $left(frac12 ight)^3{3choose3}left(frac12 ight)^4{4choose3}=frac{4}{128}$
Therefore, the probabiliy of flipping the same number of heads is: $frac{1+12+18+4}{128}=frac{35}{128}Rightarrowmathrm{(D)}$

Answer:

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