Tag: Expected Frequency

Suppose $A$ is the event of obtaining a sum of $5$ when rolling two standard dice. This means, $$A = \{(1, 4), (4, 1), (2, 3), (3, 2)\}$$so that $\text{n}(A) = 4.$ Meanwhile, the number of sample points in the experiment of rolling two standard dice is $\text{n}(S) = 6 \times 6 = 36.$ Therefore, we obtain
$$P(A) = \dfrac{\text{n}(A)}{\text{n}(S)} = \dfrac{4}{36} = \dfrac19.$$Since the experiment is conducted $n =180$ times, the expected frequency of obtaining a sum of $5$ is
$$F_h = P(A) \times n = \dfrac19 \times 180 = 20~\text{times}.$$(Answer B)

Suppose $E$ is the event of obtaining one heads and one tails when tossing two coins. This means, $$E = \{(A, G), (G, A),\}$$so that $\text{n}(E) = 2.$ Meanwhile, the number of sample points in the experiment of tossing two coins is $\text{n}(S) = 2 \times 2 = 4.$ Therefore, we obtain
$$P(E) = \dfrac{\text{n}(E)}{\text{n}(S)} = \dfrac{2}{4} = \dfrac12.$$Since the experiment is conducted $n =40$ times, the expected frequency of obtaining one heads and one tails is
$$F_h = P(E) \times n = \dfrac12 \times 40 = 20~\text{times}.$$(Answer D)

Suppose $A$ is the event that an Ace card is drawn. This means, $\text{n}(A) = 4$ because there are $4$ Ace cards in a standard deck of playing cards, namely the Ace of spades, Ace of clubs, Ace of diamonds, and Ace of hearts. Meanwhile, the number of sample points in the experiment of drawing one card is $\text{n}(S) = 52$ because there are $52$ cards in a standard deck. Therefore, we obtain
$$P(A) = \dfrac{\text{n}(A)}{\text{n}(S)} = \dfrac{4}{52} = \dfrac{1}{13}.$$Since the experiment is conducted $n =50$ times, the expected frequency of obtaining an Ace card is
$$F_h = P(A) \times n = \dfrac{1}{13} \times 50 = 3,\cdots~\text{times}.$$Notice that $3,\cdots$ lies between $3$ and $4.$ Therefore, it can be concluded that the expected frequency of obtaining an Ace card is $3$ or $4$ times.
(Answer C)

Suppose $A$ is the event that a yellow ball is drawn. This means, $\text{n}(A) = 6$ because there are $6$ yellow balls in the box. Meanwhile, the number of sample points in the experiment of drawing one ball is $\text{n}(S) = 10$ because there are $4+6=10$ balls in total in the box. Therefore, we obtain
$$P(A) = \dfrac{\text{n}(A)}{\text{n}(S)} = \dfrac{6}{10} = \dfrac35.$$Since the experiment is conducted $n =120$ times, the expected frequency of drawing a yellow ball is
$$F_h = P(A) \times n = \dfrac{3}{\cancel{5}} \times \cancelto{24}{120} = 72~\text{times}.$$Therefore, a yellow ball is expected to be drawn $72$ times.
(Answer C)

Suppose $A$ is the event that a grape-flavored candy is drawn. This means, $\text{n}(A) = 3$ because there are $3$ grape-flavored candies in the bag. Meanwhile, the number of sample points in the experiment of drawing one candy is $\text{n}(S) = 12$ because there are $12$ candies in total in the bag. Therefore, we obtain
$$P(A) = \dfrac{\text{n}(A)}{\text{n}(S)} = \dfrac{3}{12} = \dfrac14.$$Since the experiment is conducted $n =60$ times, the expected frequency of drawing a grape-flavored candy is
$$F_h = P(A) \times n = \dfrac{1}{\cancel{4}} \times \cancelto{15}{60} = 15~\text{times}.$$Therefore, a grape-flavored candy is expected to be drawn $15$ times.
(Answer C)

Suppose $A$ is the event that the wheel pointer lands on gray or purple. Since there are two intended colors, we obtain $\text{n}(A) = 2.$ Meanwhile, the number of sample points in the spinning wheel experiment is $\text{n}(S) = 4$ because the wheel is divided into $4$ equal color sections. Therefore, we obtain
$$P(A) = \dfrac{\text{n}(A)}{\text{n}(S)} = \dfrac{2}{4} = \dfrac12.$$Since the experiment is conducted $n = 100$ times, the expected frequency of the wheel pointer landing on gray or purple is
$$F_h = P(A) \times n = \dfrac{1}{\cancel{2}} \times \cancelto{50}{100} = 50~\text{times}.$$Therefore, the wheel pointer is expected to land on gray or purple $50$ times.
(Answer C)

Suppose $A$ is the event that a prize-winning raffle ticket is drawn. This means, $\text{n}(A) = 20$ because there are $20$ prize-winning raffle tickets available. Meanwhile, the number of sample points in the spinning wheel experiment is $\text{n}(S) = 50$ because there are a total of $50$ raffle tickets. Therefore, we obtain
$$P(A) = \dfrac{\text{n}(A)}{\text{n}(S)} = \dfrac{20}{50} = \dfrac25.$$Since the experiment is conducted $n = 200$ times, the expected frequency of drawing a prize-winning raffle ticket is
$$F_h = P(A) \times n = \dfrac{2}{\cancel{5}} \times \cancelto{40}{200} = 80~\text{times}.$$Therefore, the prize-winning raffle ticket is expected to be drawn $80$ times.
(Answer E)

Suppose $A$ is the event of obtaining an even number when rolling the special die. This means, $\text{n}(A) = 4$ because there are $4$ even numbers on the faces of the die, namely $2, 2, 4,$ and $6.$ Meanwhile, the number of sample points in the experiment of rolling the special die is $\text{n}(S) = 6$ because the die has $6$ faces. Therefore, we obtain
$$P(A) = \dfrac{\text{n}(A)}{\text{n}(S)} = \dfrac{4}{6} = \dfrac23.$$Since the experiment is conducted $n = 120$ times, the expected frequency of obtaining an even number is
$$F_h = P(A) \times n = \dfrac{2}{\cancel{3}} \times \cancelto{40}{120} = 80~\text{times}.$$Therefore, an even number is expected to appear $80$ times.
(Answer C)

Suppose $A$ is the event that the player wins. It is known that $P(A) = \dfrac{1}{12}$ and the experiment is conducted $n = 240$ times. Thus, the expected frequency of the player winning is
$$F_h = P(A) \times n = \dfrac{1}{\cancel{12}} \times \cancelto{20}{240} = 20~\text{times}.$$(Answer B)

Suppose $A$ is the event that the person does not obtain a mini doll. It is known that $P(A^C) = 0,\!40$ so that $P(A) = 1-0,\!40 = 0,\!60.$ Since the experiment is conducted $n = 200$ times, the expected frequency of the person not obtaining a mini doll is
$$F_h = P(A) \times n = 0,\!60 \times 200 = 120~\text{times}.$$(Answer D)

Let $E$ be the event that exactly two heads appear when the three coins are tossed. This means $\text{n}(E) = 3$ because there are three possible outcomes, namely $(H, H, T),$ $(H, T, H),$ and $(T, H, H).$ Meanwhile, the number of sample space elements for tossing three coins is $\text{n}(S) = 2 \times 2 \times 2 = 8.$ Therefore, we obtain
$$P(E) = \dfrac{\text{n}(E)}{\text{n}(S)} = \dfrac{3}{8}.$$Since the experiment is conducted $n = 120$ times, the expected frequency of exactly two heads is
$$F_h = P(E) \times n = \dfrac{3}{\cancel{8}} \times \cancelto{15}{120} = 45~\text{times}.$$Thus, exactly two heads are expected to appear $45$ times.
Note: The term “exactly two” means exactly two, no less and no more. The word “exactly” is used for emphasis.
(Answer C)

Let $A$ be the event that the first person wins against the second person in rock-paper-scissors. Let $(X, Y)$ denote the first person choosing $X$ and the second person choosing $Y.$ There are three ways for the first person to win, namely $(S, P), (R, S),$ and $(P, R),$ where $S, R, P$ represent scissors, rock, and paper respectively. Thus, $\text{n}(A) = 3.$ Meanwhile, the number of sample space elements is $\text{n}(S) = 3 \times 3 = 9.$ Therefore, we obtain
$$P(A) = \dfrac{\text{n}(A)}{\text{n}(S)} = \dfrac{3}{9} = \dfrac13.$$Since the experiment is conducted $n = 30$ times, the expected frequency is
$$F_h = P(A) \times n = \dfrac{1}{\cancel{3}} \times \cancelto{10}{30} = 10~\text{times}.$$(Answer B)

Let $A$ be the event of selecting the word “Stecu” or “Mewing”. From the image, there are $5$ words in equally sized regions: Stecu, Ohio, Skibidi, Sigma, and Mewing. Since Stecu and Mewing occupy $2$ out of $5$ regions, the probability is $P(A) = \dfrac25.$ The experiment is conducted $n = 30$ times, so the expected frequency is
$$F_h = P(A) \times n = \dfrac{2}{\cancel{5}} \times \cancelto{6}{30} = 12~\text{times}.$$(Answer C)

Let $x$ be the probability of getting an odd number. This means each of $1, 3,$ and $5$ has probability $x,$ while each of $2, 4,$ and $6$ has probability $2x.$ Since total probability is 1, we get
$$\begin{aligned} (x+x+x)+(2x+2x+2x) &= 1 \\ 9x &= 1 \\ x &= \dfrac19. \end{aligned}$$Thus, the probability of an odd number is $\dfrac19.$
Let $A$ be the event of obtaining the number $5$. Since $5$ is odd, $P(A) = \dfrac19.$ The experiment is conducted $n = 90$ times, so
$$F_h = P(A) \times n = \dfrac{1}{\cancel{9}} \times \cancelto{10}{90} = 10~\text{times}.$$(Answer B)

It is known that the number of sample space elements for rolling two dice is $\text{n}(S) = 6 \times 6 = 36.$ Meanwhile, the event of obtaining a sum of $8$ is $$A = \{(2, 6), (6, 2), (3, 5), (5, 3), (4, 4)\}$$so $\text{n}(A) = 5.$ Thus, $P(A) = \dfrac{5}{36}.$ Since the expected frequency is $15$, we get
$$\begin{aligned} F_h &= P(A) \times n \\ 15 &= \dfrac{5}{36} \times n \\ n &= \cancelto{3}{15} \times \dfrac{36}{\cancel{5}} = 108. \end{aligned}$$Thus, the value of $x$ is $\boxed{108}.$
(Answer C)

From the table, it is known that the number $4$ appears 8 times, so $n_0 = 8.$ The total number of trials is $n = 50.$ Therefore, the relative frequency of obtaining the number $4$ is
$$F_r = \dfrac{n_0}{n} = \dfrac{8}{50} = \dfrac{4}{25}.$$(Answer B)

Solution a)
Let $A$ be the event that hompimpa is repeated. All possible outcomes when hompimpa is played by three people are shown in the following table. Note: $P$ denotes white, while $H$ denotes black.
$$\begin{array}{ccc} \hline \text{Person 1} & \text{Person 2} & \text{Person 3} \\ \hline P & P & P \\ P & P & H \\ P & H & P \\ P & H & H \\ H & P & P \\ H & P & H \\ H & H & P \\ H & H & H \\ \hline \end{array}$$From the table above, the sample space of the hompimpa experiment has $\text{n}(S) = 8$ outcomes. From these $8$ outcomes, there are $2$ cases in which hompimpa is repeated, namely $(P, P, P)$ and $(H, H, H).$ Thus, $\text{n}(A) = 2.$
Therefore, we obtain
$$P(A) = \dfrac{\text{n}(A)}{\text{n}(S)} = \dfrac{2}{8} = \dfrac14.$$Thus, the probability that hompimpa is repeated is $\dfrac14.$
Solution b)
Let $B$ be the event that the first person wins against the other two. From the $8$ outcomes in the table above, there are $2$ cases in which the first person wins, namely $(H, P, P)$ and $(P, H, H).$ Thus, $\text{n}(B) = 2.$
Therefore, we obtain
$$P(B) = \dfrac{\text{n}(B)}{\text{n}(S)} = \dfrac{2}{8} = \dfrac14.$$Since the experiment is conducted $n = 80$ times, the expected frequency of event $B$ is
$$F_h = P(B) \times n = \dfrac{1}{\cancel{4}} \times \cancelto{20}{80} = 20~\text{times}.$$Thus, the expected frequency of the first person winning is $20$ times.
Solution c)
Let $C$ be the event that the first person wins against the other two, given that the first and second players always choose the same hand position. Under this condition, the sample space is given by
$$S = \{(P, P, P), (H, P, P), (P, H, H), (H, H, H)\}$$so that $\text{n}(S) = 4.$ From these $4$ outcomes, the cases in which $C$ occurs are $(H, P, P)$ and $(P, H, H),$ so $\text{n}(C) = 2.$ Therefore, we obtain
$$P(C) = \dfrac{\text{n}(C)}{\text{n}(S)} = \dfrac{2}{4} = \dfrac12.$$Since the experiment is conducted $n = 80$ times, the expected frequency of event $C$ is
$$F_h = P(C) \times n = \dfrac{1}{\cancel{2}} \times \cancelto{40}{80} = 40~\text{times}.$$Thus, the expected frequency of the first person winning against the other two, given that the first and second players always choose the same hand position, is $40$ times.