from: category_eng

Case Devision

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?

$extbf{(A) } frac{1}{8} qquad extbf{(B) } frac{5}{32} qquad extbf{(C) } frac{9}{32} qquad extbf{(D) } frac{3}{8} ...$

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?

$mathrm{(A)} frac{1}{3} qquad mathrm{(B)} frac{4}{9} qquad mathrm{(C)} frac{1}{2} qquad mathrm{(D)} frac{5}{9} qqu...$

Twelve fair dice are rolled. What is the probability that the product of the numbers on the top faces is prime?

$mathrm{(A)} left(frac{1}{12} ight)^{12} qquad mathrm{(B)} left(frac{1}{6} ight)^{12} qquad mathrm{(C)} 2left(fra...$

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?

$mathrm{(A)} {{{frac{1}{4}}}} qquad mathrm{(B)} {{{frac{2}{7}}}} qquad mathrm{(C)} {{{frac{3}{7}}}} qquad mathrm{...$

10.

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?

$mathrm{(A) } frac{29}{128} qquad mathrm{(B) } frac{23}{128} qquad mathrm{(C) } frac14 qquad mathrm{(D) } f...$

11.

For a particular peculiar pair of dice, the probabilities of rolling $1$ , $2$ , $3$ , $4$ , $5$ , and $6$ , on each die are in the ratio $1:2:3:4:5:6$ . What is the probability of rolling a total of $7$ on the two dice?

$mathrm{(A) } frac{4}{63}qquad mathrm{(B) } frac{1}{8}qquad mathrm{(C) } frac{8}{63}qquad mathrm{(D) } frac...$

12.

The wheel shown is spun twice, and the randomly determined numbers opposite the pointer are recorded. The first number is divided by $4,$ and the second number is divided by $5.$ The first remainder designates a column, and the second remainder designates a row on the checkerboard shown. What is the probability that the pair of numbers designates a shaded square?

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$extbf{(A) } frac{1}{3} qquad extbf{(B) } frac{4}{9} qquad extbf{(C) } frac{1}{2} qquad extbf{(D) } frac{5}{9} qq...$

13.

Three red beads, two white beads, and one blue bead are placed in line in random order. What is the probability that no two neighboring beads are the same color?

$mathrm{(A)} 1/12qquadmathrm{(B)} 1/10qquadmathrm{(C)} 1/6qquadmathrm{(D)} 1/3qquadmathrm{(E)} 1/2$

14.

Integers $a, b, c,$ and $d$ , not necessarily distinct, are chosen independently and at random from 0 to 2007, inclusive. What is the probability that $ad-bc$ is even?

$mathrm{(A)} frac 38qquad mathrm{(B)} frac 7{16}qquad mathrm{(C)} frac 12qquad mathrm{(D)} frac 9{16}qquad ma...$

1.

2.

3. 20/27	Complex

4. 2/9	Complex

5. 13/729	Understanding

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
$frac{1}{8}+frac{3}{32}+frac{1}{16} = frac{4}{32}+frac{3}{32}+frac{2}{32} = oxed{ extbf{(C) } frac{9}{32}}$

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. $dfrac{4}{9}+dfrac{1}{9}=oxed{mathrm{(D)} dfrac{5}{9}}$

8.

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 $dfrac{3}{6} imesleft(dfrac{1}{6}<br /> ight)^{11} imesdbinom{12}{1} = dfrac{1}{2} imesleft(dfrac{1}{6}<br /> ight)^{11} imes...$ .

9.

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 $dfrac{14}{28}=oxed{mathrm{(D)} dfrac{1}{2}}$ .

10.

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:
$frac{1+12+18+4}{128}=frac{35}{128}Rightarrowmathrm{(D)}$

11.

Let $x$ be the probability of rolling a $1$ . The probabilities of rolling a
$2$ , $3$ , $4$ , $5$ , and $6$ are $2x$ , $3x$ , $4x$ , $5x$ , and $6x$ , respectively.

The sum of the probabilities of rolling each number must equal 1, so

$x+2x+3x+4x+5x+6x=1$

$21x=1$

$x=frac{1}{21}$

So the probabilities of rolling a $1$ , $2$ , $3$ , $4$ , $5$ , and $6$ are respectively $frac{1}{21}, frac{2}{21}, frac{3}{21}, frac{4}{21}, frac{5}{21}$ , and $frac{6}{21}$ .

The possible combinations of two rolls that total $7$ are: $(1,6) ; (2,5) ; (3,4) ; (4,3) ; (5,2) ; (6,1)$

The probability of rolling a total of $7$ on the two dice is equal to the sum of the probabilities of rolling each combination.

$P = frac{1}{21}cdotfrac{6}{21}+frac{2}{21}cdotfrac{5}{21}+frac{3}{21}cdotfrac{4}{21}+frac{4}{21}cdotfrac{3}{21}+...$

12.

Solution 1

When dividing each number on the wheel by $4,$ the remainders are $1, 1, 2, 2, 3,$ and $3.$ Each column on the checkerboard is equally likely to be chosen.

When dividing each number on the wheel by $5,$ the remainders are $1, 1, 2, 2, 3,$ and $4.$

The probability that a shaded square in the $1$ st or $3$ rd row of the $1$ st or $3$ rd column is
$frac{2}{3} imes frac{3}{6} = frac{1}{3}$

The probability that a shaded square in the $2$ nd or $4$ th row of the $2$ nd column is
$frac{1}{3} imes frac{3}{6} = frac{1}{6}$

Add those two together
$frac{1}{3} + frac{1}{6} = frac{2}{6} + frac{1}{6} = frac{3}{6} = oxed{ extbf{(C)} frac{1}{2}}$

Solution 2

Alternatively, we may analyze this problem a little further.

First, we isolate the case where the rows are numbered 1 or 2. Notice that as listed before, the probability for picking a shaded square here is $frac{1}{2}$ because the column/row probabilities are the same, with the same number of shaded and non-shaded squares

Next we isolate the rows numbered 3 or 4. Note that the probability of picking the rows is same, because of our list up above. The columns, of course, still have the same probability. Because the number of shaded and non-shaded squares are equal, we have $frac{1}{2}$
Combining these we have a general probability of $oxed{ extbf{(C)} frac{1}{2}}$

13.

There are two ways to arrange the red beads.

R _ R _ R _
R _ _ R _ R

In the first, there are three ways to place a bead in the first free space, two for the second free space, and one for the third, so there are $6$ arrangements.
In the second, a white bead must be placed in the third free space, so there are two possibilities for the third space, two for the second, and one for the first. That makes $4$ arrangements. There are $6+4=10$ arrangements in total. The two cases above can be reversed, so we double $10$ to $20$ arrangements. Also, in each case, there are three ways to place the first red bead, two for the second, and one for the third, so we multiply by $6$ to get $20cdot 6 = 120$ arrangements. There are $6! = 720$ total arrangements so the answer is $120/720 = oxed{1/6}$ .

14.

The only times when $ad-bc$ is even is when $ad$ and $bc$ are of the same parity. The chance of $ad$ being odd is $frac 12 cdot frac 12 = frac 14$ , so it has a $frac 34$ probability of being even. Therefore, the probability that $ad-bc$ will be even is $left(frac 14<br /> ight)^2+left(frac 34<br /> ight)^2=frac 58 mathrm{(E)}$ .