Quiz Level-2

Hint

Let \(E\) represent the event that the sum of the faces of two dice is 7.
The possible cases for the sum to be equal to 7 are: \((1,6),(2,5),(3,4),(4,3),(5,2)\), and \((6,1)\), so event \(E\) is
\(
E=\{(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\}
\)
The probability of the event \(E[latex] is
[latex]
P(E)=6 / 36 \text { or } 1 / 6 \text {. }
\)

Hint

The sample space consists of the following six possibilities in set \(\mathrm{S}: \mathrm{S}=1,2,3,4, 5, 6\) Let \(\mathrm{E}\) be the event that the number rolled is greater than four: \(\mathrm{E}=5,6\) Therefore, the probability of \(\mathrm{E}\) is: \(\mathrm{P}(\mathrm{E})=2 / 6\) or \(1 / 3\).

Hint

We assume the marbles are \(r_1, r_2, r_3, w_1, w_2, w_3, w_4, b_1, b_2, b_3\). Let the event \(\mathrm{C}\) represent that the marble is red or blue.
The sample space \(\mathrm{S}=\left\{\mathrm{r}_1, \mathrm{r}_2, \mathrm{r}_3, \mathrm{w}_1, \mathrm{w}_2, \mathrm{w}_3, \mathrm{w}_4, \mathrm{~b}_1, \mathrm{~b}_2, \mathrm{~b}_3\right\}\).
And the event \(\mathrm{C}=\left\{\mathrm{r}_1, \mathrm{r}_2, \mathrm{r}_3, \mathrm{~b}_1, \mathrm{~b}_2, \mathrm{~b}_3\right\}\)
Therefore, the probability of \(\mathrm{C}\),
\(
\mathrm{P}(\mathrm{C})=6 / 10 \text { or } 3 / 5
\)

Hint

Since two marbles are drawn without replacement, the sample space consists of the following six possibilities.
\(
\mathrm{S}=\{(1,2),(1,3),(2,1),(2,3),(3,1),(3,2)\}
\)
Note that \((1,1),(2,2)\) and \((3,3)\) are not listed in the sample space. These outcomes are not possible when drawing without replacement, because once the first marble is drawn but not replaced into the jar, that marble is not available in the jar to be selected again on the second draw.
Let the event \(\mathrm{E}\) represent that the sum of the numbers is five. Then
\(
\mathrm{E}=\{(2,3),(3,2)\}
\)
Therefore, the probability of \(\mathrm{E}\) is
\(
\mathrm{P}(\mathrm{E})=2 / 6 \text { or } 1 / 3 \text {. }
\)

Hint

When two marbles are drawn with replacement, the sample space consists of the following nine possibilities.
\(
\mathrm{S}=\{(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3)\}
\)
Note that \((1,1),(2,2)\) and \((3,3)\) are listed in the sample space. These outcomes are possible when drawing with replacement, because once the first marble is drawn and replaced, that marble is not available in the jar to be drawn again.
Let the event \(E\) represent that the sum of the numbers is four. Then
\(
\mathrm{E}=(2,3),(3,2)
\)
Therefore, the probability of \(\mathrm{F}\) is \(\mathrm{P}(\mathrm{E})=2 / 9\)

Hint

The sample space when drawing with replacement consists of the following nine possibilities.
\(
\mathrm{S}=(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3)
\)
Let the event \(\mathrm{F}\) represent that the sum of the numbers is at least four. Then
\(
\mathrm{F}=(1,3),(3,1),(2,3),(3,2),(2,2),(3,3)
\)
Therefore, the probability of \(\mathrm{F}\) is
\(
\mathrm{P}(\mathrm{F})=6 / 9 \text { or } 2 / 3 \text {. }
\)

Hint

Let us first do an easier problem-the probability of obtaining a pair of kings and queens.
Since there are four kings, and four queens in the deck, the probability of obtaining two kings, two queens and one other card is
\(
\mathrm{P}(\text { A pair of kings and queens })=\frac{{ }^4 C_2 \times { }^4 C_2 \times { }^44 C_1}{{ }^{52} C_5}
\)
To find the probability of obtaining two pairs, we have to consider all possible pairs.
Since there are altogether 13 values, that is, aces, deuces, and so on, there are \({ }^{13} C_2\) different combinations of pairs.
\(
P(\text { Two pairs })=^{13} C_2 \cdot \frac{{ }^4 C_2 \times { }^4 C_2 \times { }^{44} C_1}{{ }^{52} C_5}=.04754
\)

Hint

There are \({ }^8 C_2\) ways of selecting 2 out of the 8 iPhones.
and \({ }^7 C_4\) ways of selecting 4 out of the 7 Android phones
But altogether there are \({ }^{15} C_6\) ways of selecting 6 out of 15 cell phones.
Therefore we have
\(
P(2 \text { iPhones and } 4 \text { Android phones })=\frac{{ }^8 C_2 \times { }^7 C_4}{{ }^{15} C_6}=\frac{(28)(35)}{5005}=\frac{980}{5005}=0.1958
\)

Hint

There are \({ }^{20} C_4\) ways of selecting 4 out of the 20 plain bagels, and \({ }^{15} C_3\) ways of selecting 3 out of the 15 poppyseed bagels, and \({ }^{18} C_2\) ways of selecting 2 out of the 18 sesame seed bagels.
But altogether there are \({ }^{53} C_9\) ways of selecting 9 out of the 53 bagels.
\(
\begin{aligned}
\mathrm{P}(4 \text { plain, } 3 \text { poppyseed, and } 2 \text { sesame seed }) & =\frac{{ }^{20} C_4 \times { }^{15} C_3 \times { }^{18} C_2}{{ }^{53} C_9} \\
& =\frac{(4845)(455)(153)}{4431613550} \\
& =0.761
\end{aligned}
\)

Hint

Let event \(\mathrm{E}\) represent that at least two people have the same birthday.
We first find the probability that no two people have the same birthday.
We analyze as follows.
Suppose there are 365 days to every year. According to the multiplication axiom, there are \(365^{25}\) possible birthdays for 25 people. Therefore, the sample space has \(365^{25}\) elements. We are interested in the probability that no two people have the same birthday. There are 365 possible choices for the first person and since the second person must have a different birthday, there are 364 choices for the second, 363 for the third, and so on. Therefore,
\(
\mathrm{P}(\text { No two have the same birthday })=\frac{365 \cdot 364 \cdot 363 \cdots 341}{365^{25}}=\frac{365 \mathrm{P} 25}{365^{25}}
\)
Since \(\mathrm{P}\) (at least two people have the same birthday) \(=1-\mathrm{P}\) (No two have the same birthday),
\(
\text { P at least two people have the same birthday })=1-\frac{365 \mathrm{P} 25}{365^{25}}=.5687
\)

Hint

It is easier to count the number of ways to roll a 4 or less. There is one way \((1,1)\) to roll a 2 , there are 2 ways to roll a \(3(1,2)\) and \((2,1)\), and 3 ways to roll a 4 \((1,3),(2,2)\), and \((3,1)\) for a total of 6 out of the \(6^2=\) 36 possible rolls, leaving 30 ways to roll greater than a \(4.30 / 36=5 / 6\).

Hint

There are \(4 !=24\) ways to arrange the digits. To be be divisible by 4 , the integer formed must end in either 12,24 , or 32 . Two of each of these makes 6 numbers divisible by \(4: 3,412 ; 4,312 ; 1,324 ; 3,124 ; 1,432\); and 4,132 . This means there are 18 out of the 24 that are not divisible by 4 or \(3 / 4\).

Hint

There are \(2^3=8\) ways to flip 2 coins. Only HHH and TTT do not show at least one tails or one heads, so there are 6 of 8 possibilities which show at least one of each,making the probability \(3 / 4\).

Hint

It is easiest to calculate the probability that the product is odd, which requires that both rolls be odd. The probability of rolling an odd number is \(1 / 2\), so the probability of rolling two odds in a row is \((1 / 2)^2=1 / 4\). Subtract this from 1 to get the probability that the product of the two rolls is even: \(3 / 4\).

Hint

We will find the probability that no two dates fall on the same day of the week and subtract this from 1 . The first date selected can be on any one the seven days. The second date must fall on one of the six remaining days, and the third date must fall on one of the remaining 5 days for a probability of: \(\frac{7}{7} \cdot \frac{6}{7} \cdot \frac{5}{7}=\frac{30}{49}\), making the probability that at least two f the dates fall on the same day of the week
\(
1-\frac{30}{49}=\frac{19}{49}
\)

Hint

It is much simpler to calculate the probability that all three are the same color. There are three blocks of each color, so the probability is the same for each: \(\frac{3}{9} \cdot \frac{2}{8} \cdot \frac{1}{7}=\frac{1}{84}\). Multiply this by 3 because there are three colors to get \(1 / 28\). The probability that no two blocks will be different colors is \(1 / 28\), so the probability that at least two will be different colored is 27/28.

Hint

We will look for the probability that \(\mathrm{A}\) and \(\mathrm{B}\) end up in the same pile. There are \({ }^9 C_3=84\) ways to choose 3 of 9 letters for a pile. There are 7 possible sets that include A, B, and one of the remaining 7 letters. This makes the probability that any pile of 3 letters will include both \(\mathrm{A}\) and \(\mathrm{B}  7 / 84=1 / 12\). There are 3 piles, so the probability that 1 of 3 piles will include both \(A\) and \(B\) is \(3(1 / 12)=3 / 12\) or \(1 / 4\). The probability that \(A\) and \(B\) will be in different piles is therefore \(1-1 / 4=\) \(3 / 4\). Alternatively, place the \(A\) first:
A _ _     _ _ _     _ _ _

This leaves 8 places for the B. 2 of those places are in the same pile as the \(A\), while 6 are in one of the other piles. \(6 / 8=3 / 4\).

Hint

Start by placing the 3. Any arrangement can be rotated so that the three is located on top. When we place the three on top, this leaves 4 blanks to fill with 4 integers for \(4 !=24\) possible ways to complete the diagram. To avoid consecutive integers, we can have a 1 and a 5 connected to the three as shown in two ways with the 2 and the 4 forced (depending on the placement of the 1 and the 5). This gives us just 2 ways to avoid consecutive integers (shown), so there are \(22 / 24=\mathbf{1 1} / 12\) arrangements with at least one connected pair of consecutive integers.

Hint

\(
{ }^3 C_{14}=\left(\begin{array}{c}
14 \\
3
\end{array}\right)=\frac{14 !}{3 !(14-3) !}=\frac{14 \cdot 13 \cdot 12}{3 \cdot 2 \cdot 1}=364
\)

Hint

Let’s denote the event of getting 3 red balls and 2 black balls by \(A\), so
\(
P(A)=\frac{n(A)}{n(S)}=\frac{\left(\begin{array}{l}
5 \\
3
\end{array}\right) \times\left(\begin{array}{l}
4 \\
2
\end{array}\right)}{\left(\begin{array}{l}
9 \\
5
\end{array}\right)}=\frac{10 \times 6}{126}=\frac{30}{63}
\)

Hint

\(
\begin{aligned}
& P(1)=P(3)=P(5)=5 P(2)=5 P(4)=5 P(6) \\
& P(1)+P(2)+P(3)+P(4)+P(5)+P(6)=1 \\
& 5 P(6)+P(6)+5 P(6)+P(6)+5 P(6)+P(6)=1 \\
& 18 P(6)=1 \\
& P(6)=1 / 18
\end{aligned}
\)

Hint

Since there are the same amount of total balls in Alice’s bag as in Bob’s bag, and there is an equal chance of each ball being selected, the color of the ball that Alice puts in Bob’s bag doesn’t matter. Without loss of generality, let the ball Alice puts in Bob’s bag be red.
For both bags to have the same contents, Bob must select one of the 2 red balls out of the 6 balls in his bag.
So the desired probability is \(\frac{2}{6}=\frac{1}{3}\).

Hint

Let \(x\) be the probability of rolling a 1 . The probabilities of rolling a \(2,3,4,5\), and 6 are \(2 x, 3 x, 4 x, 5 x\), and \(6 x\), respectively. The sum of the probabilities of rolling each number must equal 1, so
\(
\begin{aligned}
& x+2 x+3 x+4 x+5 x+6 x=1 \\
& 21 x=1 \\
& x=\frac{1}{21}
\end{aligned}
\)
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 \(P\) of rolling a total of 7 on the two dice is equal to the sum of the probabilities of rolling each combination.
\(
P=\frac{1}{21} \cdot \frac{6}{21}+\frac{2}{21} \cdot \frac{5}{21}+\frac{3}{21} \cdot \frac{4}{21}+\frac{4}{21} \cdot \frac{3}{21}+\frac{5}{21} \cdot \frac{2}{21}+\frac{6}{21} \cdot \frac{1}{21}=\frac{8}{63}
\)

Hint

Total no. of ways \(={ }^{52} \mathrm{C}_2\)
Case I: Both are diamonds \(={ }^{13} \mathrm{C}_2\)
Case II: Both are kings \(={ }^4 \mathrm{C}_2\)
\(P\) (both are diamonds or both are kings) \(=\left({ }^{13} \mathrm{C}_2+{ }^4 \mathrm{C}_2\right) /{ }^{52} \mathrm{C}_2\)

Hint

Probability of the problem getting solved \(=1-\) (Probability of none of them solving the problem)
\(
P(P)=\frac{2}{7} \Rightarrow P(\bar{P})=1-\frac{2}{7}=\frac{5}{7}, P(Q)=\frac{4}{7} \Rightarrow P(\bar{Q})=1-\frac{4}{7}=\frac{3}{7}, P(R)=\frac{4}{9} \Rightarrow P(\bar{R})=1-\frac{4}{9}=\frac{5}{9}
\)
Probability of problem getting solved \(=1-(5 / 7) \times(3 / 7) \times(5 / 9)=(122 / 147)\)

Hint

A leap year can have 52 Sundays or 53 Sundays. In a leap year, there are 366 days out of which there are 52 complete weeks & remaining 2 days. Now, these two days can be (Sat, Sun) (Sun, Mon) (Mon, Tue) (Tue, Wed) (Wed, Thur) (Thur, Friday) (Friday, Sat).
So there are total 7 cases out of which (Sat, Sun) (Sun, Mon) are two favorable cases. So, P (53 Sundays) \(=2 / 7\)
Now, \(\mathrm{P}(52\) Sundays \()+\mathrm{P}(53\) Sundays \()=1\)
So, \(P(52\) Sundays \()=1-P(53\) Sundays \()=1-(2 / 7)=(5 / 7)\)

Hint

Let E1, E2, E3 and A are the events defined as follows.
\(\mathrm{E} 1\) = First bag is chosen
E2 = Second bag is chosen
E3 \(=\) Third bag is chosen
\(A=\) Ball drawn is red
Since there are three bags and one of the bags is chosen at random, so \(P(E 1)=P(E 2)=P(E 3)=1 / 3\)
If \(\mathrm{E} 1\) has already occurred, then first bag has been chosen which contains 3 red and 7 black balls. The probability of drawing 1 red ball from it is \(3 / 10\). So, \(P\left(A / E_1\right)=3 / 10\), similarly \(P\left(A / E_2\right)=8 / 10\), and \(P\left(A / E_3\right)=\) 4/10. We are required to find \(P\left(E_3 / A\right)\) i.e. given that the ball drawn is red, what is the probability that the ball is drawn from the third bag by Baye’s rule
\(
=\frac{\frac{1}{3} \times \frac{4}{10}}{\frac{1}{3} \times \frac{3}{10}+\frac{1}{3} \times \frac{8}{10}+\frac{1}{3} \times \frac{4}{10}}=\frac{4}{15} .
\)

Hint

There are \({ }^{13} C_5=1,287\) ways to choose 5 cards from a set of 13. If we have \(\mathrm{KQJ}\) in the set, we must choose 2 of the remaining 10 cards to complete the set. There are \({ }^{10} C_2=45\) ways to complete the set, making the probability \(45 / 1,287=5 / 143\).

Hint

There are \({ }^{12} C_3=220\) ways to select three students from a group of 12. If Adam is selected, there are 10 students left (excluding Billy) and we need to choose two to pair with Adam. \({ }^{10} \mathrm{C}_2=45\) pairs can be selected, meaning there are 45 groups of 3 students which include Adam but not Billy out of the 220 possible groups. \(45 / 220=9 / 44\).

Hint

There are \({ }^{20} C_3=1,140\) sets of 3 students who can be chosen from a group of 20 . We are looking to form a group with 1 vegetarian and 2 non-vegetarians. There are \({ }^3 C_1=3\) ways to choose the vegetarian and \({ }^{17} \mathrm{C}_2=136\) ways to choose the non-vegetarians, for a total of \(3 \cdot 136=408\) ways to form a group with one vegetarian and two non-vegetarians, which gives us a probability of \(408 / 1,140=34 / 95\). Alternatively, The probability of selecting vegetarian, non-vegetarian, non-vegetarian in that order is: \(\frac{3}{20} \cdot \frac{17}{19} \cdot \frac{16}{18}=\frac{34}{285}\), and we can arrange these picks three different ways (the veg. can be \(1^{\text {st }}, 2^{\text {nd }}\), or \(3^{\text {rd }}\) ), which gives us: \(3 \cdot \frac{34}{285}=\frac{34}{95}\)

Hint

The probability of the point being inside the circle is the ratio of the area of the circle to the area of the square. If we suppose that the circle has radius \(r\), then the square must have side \(2 r\). The area of the circle is \(\pi r^2\) and the area of the square is \((2 r)^2=(4 r)^2\), so the proportion of the areas is \(\left(\pi r^2\right) / (4 r)^2=\pi / 4\).

Hint

If \(x\) is chosen at random from the set \((4,6,7,9,11)\) and \(y\) is chosen at random from the set \((12,13,15,17)\) then what is the probability that \(x y\) is odd?
Here we have 5 possible choices for \(x\) and 4 possible choices for \(y\), giving us \(5 \times 4=20\) possible outcomes.
We know that odd times odd = odd; even times even = even; and even times odd = even. Thus we need all of the outcomes where \(x\) and \(y\) are odd. We have 3 possibilities of odd numbers for \(x\), and 3 possibilities of odd numbers for \(y\), so we will have 9 outcomes of our total 20 outcomes where \(x y\) is odd, giving us a probability of \(9 / 20\).

Hint

There are 18 marbles in total. One of them is removed so now there are 17 marbles. This is our denominator. All of the original white marbles are still in the bag so there is a 4 out of 17 or \(4 / 17\) chance that the next marble taken out of the bag will be white.

Hint

Michael can toss at least one head, or he can toss zero heads. The sum of these two probabilities must equal one, because they represent all of the ways that Michael could toss the coins. He could either toss at least on head, or he could toss no heads at all.
Probability of tossing at least one head + probability of tossing no heads \(=1\)
The probability of tossing no heads is only possible with the combination TTT. The probability of tossing three tails is equal to \((1 / 2)(1 / 2)\) \((1 / 2)=1 / 8\)
Probability of tossing at least one head \(+1 / 8=1\)
Probability of tossing at least one head \(=1-1 / 8=7 / 8\).

Hint

Probability of each event \(=(\#\) green marbles \(+\) \# blue marbles \() /\) Total # of Marbles
\(
\mathrm{P} 1=(15+25) / 125=40 / 125
\)
Second event assumes a blue or green was chosen for first event so there is one fewer marble on top and also one fewer marble in the total number of marbles.
\(
\mathrm{P} 2=(14+25) / 124=39 / 124
\)
Third event assumes a blue or green was chosen for first and second events so there are two fewer marbles on top and also two fewer marbles in the total number of marbles.
\(
\mathrm{P} 3=(13+25) / 124=38 / 123
\)
Probability for multiple events \(=\mathrm{P} 1 \times \mathrm{P} 2 \times \mathrm{P} 3\)
\(
\begin{aligned}
& (40 / 125) \times (39 / 124) \times (38 / 123) \\
& (40 \times 39 \times 38) /(125 \times 124 \times 123)=59280 / 1906500=0.031
\end{aligned}
\)

Hint

First, observe that multiples of 5 are all those numbers that have a 0 or a 5 as their final digit \((5,10,15,20,25\), etc.), so the question is asking how many 4-digit integers under 7000 end in a 0 or a 5 . Second, notice that the smallest 4-digit integer is 1,000. So, to rephrase the question, we want to find how many integers that are between 1000 and 6999 end in a 0 or a 5.

Writing the 4-digit integer as \(W X Y Z\), we see that there are six possibilties for \(W(1,2,3,4,5\), and 6\()\), ten possibilities for \(X(0-9)\), ten possibilities for \(Y(0-9)\), and two possibilities for \(Z(0\) and 5). So there are \(6 \times 10 \times 10 \times 2=1200\) total integers between 1000 and 6999 that end in a 0 or a 5. Therefore, there are 1200 four-digit integers that are multiples of 5 and less than 7,000.

Hint

First, find the amount of time that the light is yellow, which is \(50-20-25=5\) seconds out of a 50 second cycle. The period of 100 seconds is irrelevant because the probability of getting a yellow light would be the same for any complete period or number of complete periods. So the probability of getting a yellow light would be \(5 / 50=1 / 10\).

Hint

We can find the individual probabilities of these three events occuring first.
\(P(\) rolling a 6\()=1 / 6\)
\(P(\) head \()=1 / 2\)
\(P(\) spade \()=13 / 52=1 / 4\)
Now, to find the probability of rolling a 6 AND flipping a head AND drawing a spade, we must multiply the individual probabilities. So the answer is \(1 / 6 \times 1 / 2 \times 1 / 4=1 / 48\).

Hint

There are a total of \(7+5+3+6=21\) marbles. The probability of picking a yellow marble is \(7 / 21=1 / 3\). Then we put it back and choose a blue marble with probability \(3 / 21=1 / 7\). We do NOT put this blue marble back, but then we grab for a white. The probability of picking a white is now \(6 / 20=3 / 10\), because now we are choosing from 20 marbles instead of 21 . So putting it together, the probability of choosing a yellow marble, replacing it and then choosing a blue and a white, is \(1 / 3 \times 1 / 7 \times 3 / 10=1 / 70\).

Hint

\(
\text { There are } 5 \text { black marbles worth } \$ 1000 \text { dollars out of the total of } 200 \text { marbles. } 5 \div 200=.025 \text { or } 2.5 \%
\)