Tag: Event

An event with probability $0$ means that the event is impossible to occur.

1. Obtaining a die face of $7$ from rolling a die is impossible because the highest number on a die is $6$.
2. The birth of a baby boy is a common/possible event.
3. Drawing a card numbered $11$ from a standard deck of cards is impossible because the numbered cards only go up to $10$.
4. Death is a certain event experienced by every human being (has probability $1$).
5. The sun rising every morning is a certain event (has probability $1$).
6. The appearance of fire in the depths of the ocean is impossible because fire cannot burn underwater.
7. A cat being able to speak Indonesian is an impossible event.

Based on the explanation above, it can be concluded that the events with probability $0$ are 1, 3, 6, and 7.
(Answer A)

Suppose $M$ is the event of obtaining $2$ heads ($H$) and $1$ tail $(T),$ then
$M = \{(H, H, T), (H, T, H), (T, H, H)\}$
with $n(M) = 3.$
The number of sample space members for $3$ coins, each having $2$ sides, is $n(S) = 2 \times 2 \times 2 = 8.$
Therefore, the probability is $\boxed{p(M) = \dfrac{n(M)} {n(S)} = \dfrac{3}{8}}.$
(Answer B)

Suppose $M$ is the event of obtaining $1$ head ($H$), which means the other coins show tails ($T$), so
$M = \{(H, T, T), (T, H, T), (T, T, H)\}$ with $n(M) = 3.$
The number of sample space members for $3$ coins, each having $2$ sides, is $n(S) = 2 \times 2 \times 2 = 8.$
Therefore, the probability is $\boxed{p(M) = \dfrac{n(M)} {n(S)} = \dfrac{3}{8}}.$
(Answer A)

Suppose $M$ is the event of obtaining at least $1$ head $(H)$ so that
$$\begin{aligned} & M = \{(H, T, T), (T, H, T), (T, T, H), \\ & (H, H, T), (H, T, H), (T, H, H), (H, H, H)\} \end{aligned}$$with $T$ representing tails and $n(M) = 7.$
The number of sample space members for $3$ coins, each having $2$ sides, is $n(S) = 2 \times 2 \times 2 = 8.$
Therefore, the probability is $p(M) = \dfrac{n(M)} {n(S)} = \dfrac{7}{8}.$
The expected frequency of obtaining at least $1$ head from the coin tosses is $\boxed{p(M) \times n = \dfrac{7}{8} \times 80 = 70~\text{times}}.$
(Answer A)

Suppose $A$ is the event that the sum of the dice is $4$ so that
$A = \{(1, 3), (3, 1), (2, 2)\}$ with $n(A) = 3.$
The number of sample space members for $2$ dice, each having $6$ sides, is $n(S) = 6 \times 6 \times = 36.$
Therefore, the probability is $\boxed{p(A) = \dfrac{n(A)} {n(S)} = \dfrac{3}{36}= \dfrac{1}{12}}.$
(Answer A)

A sum greater than $7$ means it can be $8, 9, 10, 11$, or $12$.
Suppose $A$ is the event that the sum of the dice is $8$ so that
$A = \{(2, 6), (6, 2), (3, 5), (5, 3), (4, 4)\}$ with $n(A) = 5.$
Suppose $B$ is the event that the sum of the dice is $9$ so that
$B = \{(3, 6), (6, 3), (4, 5), (5, 4)\}$ with $n(B) = 4.$
Suppose $C$ is the event that the sum of the dice is $10$ so that
$C = \{(4, 6), (6, 4), (5, 5)\}$ with $n(C) = 3.$
Suppose $D$ is the event that the sum of the dice is $11$ so that
$D = \{(5, 6), (6, 5)\}$ with $n(D) = 2.$
Suppose $E$ is the event that the sum of the dice is $12$ so that
$E = \{(6, 6)\}$ with $n(E) = 1.$
The number of sample space members for $2$ dice, each having $6$ sides, is $n(S) = 6 \times 6 = 36.$
Therefore, the probability is
$$\begin{aligned} & p(A \cup B \cup C \cup D \cup E) \\ & = \dfrac{n(A) + n(B) + n(C) + n(D) + n(E)} {n(S)} \\ & = \dfrac{5+4+3+2+1}{36} \\ & = \dfrac{15}{36} = \dfrac{5}{12}. \end{aligned}$$(Answer B)

A sum less than $5$ means it can be $2, 3$, or $4.$
Suppose $A$ is the event that the sum of the dice is $2$ so that $A = \{(1, 1)\}$ with $n(A) = 1.$
Suppose $B$ is the event that the sum of the dice is $3$ so that
$B = \{(1, 2), (2, 1)\}$ with $n(B) = 2.$
Suppose $C$ is the event that the sum of the dice is $4$ so that
$C = \{(1, 3), (3, 1), (2, 2)\}$ with $n(C) = 3.$
The number of sample space members for 2 dice, each having $6$ sides, is $n(S) = 6 \times 6 \times = 36.$
Therefore, the probability is
$$\begin{aligned} p(A \cup B \cup C) & = \dfrac{n(A) + n(B) + n(C)} {n(S)} \\ & = \dfrac{1+2+3}{36} \\ & = \dfrac{6}{36} = \dfrac{1}{6} \end{aligned}$$(Answer A)

Known:
Total number of eggs = $200$
Number of broken eggs = $10$
Number of unbroken eggs = $190.$
The probability of selecting an unbroken egg is
$$\boxed{\dfrac{\text{Number of unbroken eggs}} {\text{Total number of eggs}} =\dfrac{190}{200} =\dfrac{19}{20}}$$(Answer A)

Suppose $B$ represents the event of selecting a blue marble from the bag.
The number of blue marbles in the bag is
$n(B) = 60 -8 -12 -16 = 24.$
The total number of marbles is $n(S) = 60.$
Therefore, the probability of selecting a blue marble is
$\boxed{p(B) = \dfrac{n(B)} {n(S)} = \dfrac{24}{60} = \dfrac25}.$
(Answer C)

Suppose $M$ represents the event of selecting a red marble from the bag.
The number of red marbles in the bag is $n(M) = 32.$
The total number of marbles is
$n(S) = 30 + 18 + 32 = 80.$
Therefore, the probability of selecting a red marble is
$\boxed{p(M) = \dfrac{n(M)} {n(S)} = \dfrac{32}{80} = \dfrac{4}{10} = 0,\!40}$
(Answer B)

There are $2$ red faces.
A cube has $6$ faces.
The probability that the top face is red is $\boxed{\dfrac26 = \dfrac13}$
(Answer C)

Suppose $D$ represents the event that a person gets a door prize.
The number of door prizes provided is $n(D) = 9.$
The number of attendees is $n(S) = 180.$
Therefore, the probability that a person gets a door prize is $\boxed{p(D) = \dfrac{n(D)} {n(S)}= \dfrac{9}{180} = 0,\!05}$
(Answer C)

The possible outcomes when rolling a die are $\{1,2,3,4,5,6\},$ where $2, 3, 5$ (a total of $3$ numbers) are prime numbers.
Suppose the event of obtaining a prime number is denoted by $A$.
Thus, the probability of obtaining a prime number is $p(A) = \dfrac{3}{6} = \dfrac12.$
The expected frequency of obtaining a prime number from $n = 120$ rolls is
$\begin{aligned} f_h & = p(A) \times n \\ & = \dfrac12 \times 120 = 60. \end{aligned}$
Therefore, the expected frequency of obtaining a prime number is $\boxed{60~\text{times}}$
(Answer D)

The letter N appears 2 times in the word INDONESIA. The word consists of 9 letters. Therefore, the probability of selecting the letter N is $\dfrac29.$
(Answer B)

In a bridge deck (playing cards), there are $52$ cards.
The numbered cards start from $1$ (ace card) to $10$, each consisting of $4$ suits, namely heart ♥, spade ♠, diamond ♦, and club ♣.
Since there are $5$ even numbers, namely $2, 4, 6, 8,$ and $10$, and each has $4$ suits, the total number of even-numbered cards is $4 \times 5 = 20.$
Suppose the event of drawing an even-numbered card is denoted by $A$, then $\boxed{P(A) = \dfrac{20}{52} = \dfrac{5}{13}}$
(Answer D)

The probability of having a boy is equal to the probability of having a girl, namely $\dfrac12$.
The probability that both children are girls ($2$ events) is
$\underbrace{\dfrac12}_{\text{girl}} \times \underbrace{\dfrac12}_{\text{girl}} = \dfrac14.$
(Answer A)

Balls numbered $1, 2, 3$: red.
Balls numbered $4, 5, 6, 7, 8$: yellow.
Balls numbered $9, 10, 11, 12$: green.
On the first draw, a red ball with an even number appears, meaning ball number $2$ has been taken.
On the second draw, a green ball with a prime number appears, meaning ball number $11$ has been taken.
The remaining odd-numbered balls are: $1, 3, 5, 7, 9$ (there are $5$ balls).
The total number of balls remaining is $12 -2 = 10.$
Therefore, the probability of drawing an odd-numbered ball on the third draw is $\boxed{\dfrac{5}{10} = 50\%}$
(Answer C)

Balls numbered $1, 2, 3, 4, 5, 6, 7, 8$: red.
Balls numbered $9, 10, 11, 12, 13, 14$: blue.
Balls numbered $15, 16, 17, 18$: green.
The first draw produces ball number $7$.
The second draw produces ball number $13$.
The remaining even-numbered green balls are: $16, 18$ (there are $2$ balls).
The total number of balls remaining is $18 -2 = 16.$
Therefore, the probability that the third ball drawn is an even-numbered ball is $\boxed{\dfrac{2}{16}}$
(Answer A)

There are $6$ red candies.
The total number of candies is $6 + 5 + 3 + 3 + 2 + 4 + 2 + 5 = 30$ candies.
Therefore, the probability that Dilla takes a red candy is $\boxed{\dfrac{6}{30} = \dfrac{1}{5} = 20\%}$
(Answer B)

The quantity of red candies in degree units is
$$\begin{aligned} & 360^{\circ} -(18+36+108+36+18+90)^{\circ} \\ & = 360^{\circ} -306^{\circ} = 54^{\circ} \end{aligned}$$The number of red candies in the bag is
$\text{n}(\text{red}) = \dfrac{54^{\circ}} {\cancelto{9}{360}^{\circ}} \times \cancel{40} = 6.$
The probability of drawing a red candy is
$p(\text{red}) = \dfrac{\text{n(red)}} {\text{n} (S)} = \dfrac{6}{40} = 15\%.$
(Answer B)

Suppose $A$ is the event that a female student is chosen as class leader, with $n(A) = 24$ and $n(S) = 40$ so that $p(A) = \dfrac{n(A)} {n(S)} = \dfrac{24}{40} = \dfrac35.$
Suppose $B$ is the event that a male student is chosen as vice class leader, with $n(B) = 16$ and $n(S) = 40 -1 = 39$ (reduced by $1$ because one female student has already been chosen as class leader) so that $p(B) = \dfrac{n(B)} {n(S)} = \dfrac{16}{39}.$
Thus, the probability that a female student is chosen as class leader and a male student is chosen as vice class leader is
$\boxed{\begin{aligned} p(A \cap B) & = \dfrac{n(A)} {n(S)} \times \dfrac{n(B)} {n(S)} \\ & = \dfrac{3}{5} \times \dfrac{16}{39} = \dfrac{16}{65} \end{aligned}}$
(Answer C)

Observe the following table.

The blue-colored cells in the table represent events where both of them shop on consecutive days. From the table above, there are $8$ blue cells, while the total number of cells is $25$. Therefore, the probability is $\boxed{\dfrac{8}{25} = 0,\!32}$
(Answer C)

Observe the following table.
The pairs of consecutive numbers are
$$(1, 2), (2, 3), (3, 4), (4, 5), (5, 6), (6, 7), (7, 8).$$or vice versa. Since two balls are drawn simultaneously, $(1, 2)$ is considered the same as $(2, 1).$ Therefore, there are only $7$ possible cases.
The total number of sample space members is $7 + 6 + 5 + 4 + 3 + 2 + 1 = 28$ (the number of green-colored cells in the table above).
Therefore, the probability of drawing two balls with consecutive numbers is $\boxed{\dfrac{7}{28} = \dfrac14}$
(Answer D)

Alternative 1: Observe the following table.
Pairs of odd-numbered balls are marked by orange-colored cells in the table above, totaling $5 + 4 + 3 + 2 + 1 = 15.$
Since two balls are drawn simultaneously, $(1, 3)$ is considered the same as $(3, 1).$
The total number of sample space members is
$\begin{aligned} 11 + 10 + & 9 + 8 + 7 + 6 + 5 + 4 +\\ & 3 + 2 + 1 = 66. \end{aligned}$
(see the number of green-colored cells in the table above).
Therefore, the probability of drawing two odd-numbered balls is $\boxed{\dfrac{15}{66}}$
Alternative 2:
From the numbers $1$ to $12$, there are $6$ odd numbers. The probability of drawing one odd number from those twelve numbers is $P(A) = \dfrac{6}{12}.$ The probability of drawing another odd number from the remaining eleven numbers is $P(B) = \dfrac{5}{11}.$ Thus,
$P(A \cap B) = \dfrac{6}{12} \times \dfrac{5}{11} = \dfrac{30}{132} = \dfrac{15}{66}.$
(Answer C)

Convert each fraction so that the denominator becomes $24.$
The blue sections are $\dfrac{7}{24}.$
The purple sections are $\dfrac18 = \dfrac{3}{24}.$
The yellow sections are $\dfrac{5}{12} = \dfrac{10}{24}.$
The red sections are the remainder, namely $\dfrac{24-7-3-10}{24} = \dfrac{4}{24}.$
From this, it is known that purple is the hardest color to obtain because it has the fewest sections, namely $3$ out of $24$ total sections. In other words, the probability that the arrow points to purple is the smallest.
(Answer B)

Suppose the probability of obtaining each die face other than $1$ is $x,$ so the probability of obtaining face $1$ is $\dfrac15x.$ Since the total probability of all events is $1,$ we obtain
$$\begin{aligned} \dfrac15x + x + x + x + x + x & = 1 \\ \dfrac{26}{5}x & = 1 \\ x & = \dfrac{5}{26}.\end{aligned}$$The probability of obtaining an even number $(2, 4, 6)$ is given by
$$\begin{aligned} P(\text{even}) & = \dfrac{5}{26}+ \dfrac{5}{26}+\dfrac{5}{26} \\ & = \dfrac{15}{26} \end{aligned}$$(Answer C)

The number of children who only like mathematics is $n(M) = 25-15 = 10.$
The number of children who only like physics is $n(M) = 20-15 = 5.$
The total number of children in the group is $$n(S) = (25-15)+(20-15)+15 = 30.$$Answer a)
The number of children who like both subjects is $15$ people. The probability of calling them is $\dfrac{15}{30} = \dfrac12.$
Answer b)
The number of children who only like mathematics is $n(M) = 25-15 = 10.$
The probability of calling a child who only likes mathematics is $\dfrac{10}{30} = \dfrac13.$

Two events are said to be independent if one event does not affect the possibility of the occurrence of the other event. When we roll two dice, the appearance or non-appearance of the number $4$ on the first die does not affect the possibility of the appearance of the number $4$ on the second die. In this case, the probability of getting the number $4$ on both dice is equal to the product of the probabilities of getting the number $4$ on each die, namely $P(A \cap B) = P(A) \times P(B)$.
It can be concluded that $A$ and $B$ are independent events.

The number of bridge cards is $52$, while King cards consist of $4$ cards, namely K spade ♠, K heart ♥, K diamond ♦, and K club ♣. Therefore, the probability of drawing a King card from the $52$ cards is $\color{blue}{\dfrac{4}{52} = \dfrac{1}{13}}.$
The expected frequency of drawing a King card from $\color{red}{260}$ draws is
$f_h = \color{blue}{\dfrac{1}{13}} \times \color{red}{260} = 20.$
This means that from $260$ draws, we are expected to get $20$ King cards.

The probability of a successful rocket launch is $\dfrac{9}{10}.$
Since there are $50$ rocket launches, the number of rockets expected to launch successfully is $\dfrac{9}{\cancel{10}} \times \cancelto{5}{50} = 45$ units.

Answer a)
The sample space of tossing a coin (having $2$ sides: heads and tails) is $\{H, T\}$.
Answer b)
The sample space of the spinner with $3$ colors, namely red, green, and blue, is $\{\text{red}, \text{green}, \text{blue}\}.$
Answer c)
The probability of getting the number side on the coin toss is $\dfrac12.$
The probability of the spinner pointing to blue is $\dfrac13.$
Thus, the probability that both events occur is
$P(A) = \dfrac12 \times \dfrac13 = \dfrac16.$

It is known that $\text{n}(S) = 20.$
Answer a)
Square numbers are numbers resulting from squaring.
It is known that $A = \{1, 4, 9, 16\}$ so that $\text{n}(A) = 4$.
The probability of drawing a card numbered with a square number is
$P(A) = \dfrac{\text{n}(A)}{\text{n}(S)} = \dfrac{4}{20} = \dfrac15.$
Answer b)
Cube numbers are numbers resulting from cubing.
It is known that $B = \{1, 8\}$ so that $\text{n}(B) = 2$.
The probability of drawing a card numbered with a cube number is
$P(B) = \dfrac{\text{n}(B)}{\text{n}(S)} = \dfrac{2}{20} = \dfrac{1}{10}$.
Answer c)
The chosen card is numbered less than $10$ and even.
It is known that $C = \{2, 4, 6, 8\}$ so that $\text{n}(C) = 4$.
The probability of drawing a card numbered with a square number is
$P(C) = \dfrac{\text{n}(C)}{\text{n}(S)} = \dfrac{4}{20} = \dfrac15$.
Answer d)
The chosen card is numbered greater than $14$ and odd.
It is known that $D = \{15, 17, 19\}$ so that $\text{n}(D) = 3.$
The probability of drawing a card numbered with a square number is
$P(D) = \dfrac{\text{n}(D)}{\text{n}(S)} = \dfrac{3}{20}.$

The total number of cards in one deck is $\text{n}(S) = 52.$
Answer a)
The number of Queen cards is $\text{n}(\text{Q}) = 4.$
The probability of drawing a Queen card is
$P(\text{Q}) = \dfrac{\text{n}(\text{Q})}{\text{n}(S)} = \dfrac{4}{52} = \dfrac{1}{13}.$
Answer b)
The number of cards numbered $8$ or $9$ is $\text{n}(8~\text{or}~9) = 4+4 = 8$.
The probability of drawing a card numbered $8$ or $9$ is
$$\begin{aligned} P(8~\text{or}~9) & = \dfrac{\text{n}(8~\text{or}~9)}{\text{n}(S)} \\ & = \dfrac{8}{52} = \dfrac{2}{13}. \end{aligned}$$Answer c)
The number of even-numbered cards $(2, 4, 6, 8, 10)$ is $\text{n}(\text{even}) = 5 \times 4 = 20.$
The probability of drawing an even-numbered card is
$P(\text{even}) = \dfrac{\text{n}(\text{even})}{\text{n}(S)} = \dfrac{20}{52} = \dfrac{5}{13}.$
Answer d)
The number of cards numbered $7$ is $\text{n}(7) = 4.$
The number of $♠$ cards is $\text{n}(♠) = 1 \times 13 = 13.$
Note that there is $1$ card that is both numbered $7$ and $♠$, so $\text{n}(7~\text{or spade}) = 4+13-1 = 16.$
The probability of drawing a card numbered $7$ or $♠$ is
$\begin{aligned} P(7~\text{or}~♠) & = \dfrac{\text{n}(7~\text{or}~♠)}{\text{n}(S)} \\ & = \dfrac{16}{52} = \dfrac{4}{13}. \end{aligned}$

Answer a)
Total number of cards = $52$.
Number of face cards = $4 \times 3 = 12.$
The probability of drawing a non-face card (Jack, Queen, King) on the first draw is
$P(A) = \dfrac{52-12}{52} = \dfrac{40}{52} = \dfrac{10}{13}.$
The card is returned, so the total number of cards remains $52$.
The probability of drawing a face card on the second draw is
$P(B) = \dfrac{12}{52} = \dfrac{3}{13}.$
Therefore, the probability of both events occurring is
$\begin{aligned} P(A \cap B) & = P(A) \times P(B) \\ & = \dfrac{10}{13} \times \dfrac{3}{13} = \dfrac{30}{169}. \end{aligned}$
Answer b)
Total number of cards = $52$.
Number of face cards = $4 \times 3 = 12.$
The probability of drawing a non-face card (Jack, Queen, King) on the first draw is
$P(A) = \dfrac{52-12}{52} = \dfrac{40}{52} = \dfrac{10}{13}.$
The card is not returned, so the number of cards becomes $51$, where $1$ non-face card has already been drawn.
The probability of drawing a face card on the second draw is
$P(B) = \dfrac{12}{51} = \dfrac{4}{17}.$
Therefore, the probability of both events occurring is
$\begin{aligned} P(A \cap B) & = P(A) \times P(B) \\ & = \dfrac{10}{13} \times \dfrac{4}{17} = \dfrac{40}{221}. \end{aligned}$