Explanation for the Arithmetic Quiz 09

This is the detailed explanation for the Arithmetic Quiz 09. Each question’s solution is broken down step-by-step to help you understand the core concepts.

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1. Three unbiased coins are tossed. What is the probability of getting at most two heads?

Answer: 7/8

The sample space for tossing three coins is: {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT} There are a total of 8 possible outcomes. “At most two heads” means getting 0, 1, or 2 heads.

• 0 heads: {TTT} (1 outcome)
• 1 head: {HTT, THT, TTH} (3 outcomes)
• 2 heads: {HHT, HTH, THH} (3 outcomes) The total number of favorable outcomes is $1+3+3=7$. The probability is (Favorable Outcomes) / (Total Outcomes) = 7/8.

Alternatively, you can calculate the probability of the opposite event (getting exactly 3 heads) and subtract it from 1.

• The outcome for 3 heads is {HHH} (1 outcome).
• The probability of 3 heads is 1/8.
• The probability of at most two heads is $1-1/8=7/8$.

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2. Two dice are tossed. The probability that the total score is a prime number is:

Answer: 5/12

The total number of outcomes when two dice are tossed is $6\times 6=36$. The possible prime number sums are 2, 3, 5, 7, and 11.

• Sum of 2: (1, 1) – 1 outcome
• Sum of 3: (1, 2), (2, 1) – 2 outcomes
• Sum of 5: (1, 4), (2, 3), (3, 2), (4, 1) – 4 outcomes
• Sum of 7: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) – 6 outcomes
• Sum of 11: (5, 6), (6, 5) – 2 outcomes The total number of favorable outcomes is $1+2+4+6+2=15$. The probability is 15/36, which simplifies to 5/12.

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3. In a class, there are 15 boys and 10 girls. Three students are selected at random. The probability that 1 girl and 2 boys are selected, is:

Answer: 21/46

• Total number of students: $15+10=25$.
• Total number of ways to select 3 students: This is a combination problem, as the order of selection doesn’t matter. Total outcomes = $C(25,3)=(25\times 24\times 23)/(3\times 2\times 1)=2300$.
• Number of ways to select 1 girl: $C(10,1)=10$.
• Number of ways to select 2 boys: $C(15,2)=(15\times 14)/(2\times 1)=105$.
• Number of favorable outcomes (1 girl AND 2 boys) = $C(10,1)\times C(15,2)=10\times 105=1050$.
• Probability = (Favorable Outcomes) / (Total Outcomes) = $1050/2300=105/230=21/46$.

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4. What is the probability of getting a sum 9 from two throws of a dice?

Answer: 1/9

• Total outcomes when throwing two dice = $6\times 6=36$.
• Favorable outcomes (pairs that sum to 9): (3, 6), (4, 5), (5, 4), (6, 3).
• There are 4 favorable outcomes.
• Probability = (Favorable Outcomes) / (Total Outcomes) = $4/36=\ast \ast 1/9\ast \ast$.

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5. In a lottery, there are 10 prizes and 25 blanks. A lottery is drawn at random. What is the probability of getting a prize?

Answer: 2/7

• Total number of lottery tickets = $10(\text{prizes})+25(\text{blanks})=35$.
• Number of prize tickets = 10.
• Probability of getting a prize = (Number of Prize Tickets) / (Total Tickets) = $10/35=\ast \ast 2/7\ast \ast$.

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6. Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5?

Answer: 2/5

• Total outcomes: 20.
• Multiples of 3: {3, 6, 9, 12, 15, 18} – 6 numbers.
• Multiples of 5: {5, 10, 15, 20} – 4 numbers.
• Multiples of both 3 and 5 (i.e., multiples of 15): {15} – 1 number.
• Total favorable outcomes (Multiples of 3 OR 5) = (Multiples of 3) + (Multiples of 5) – (Multiples of both) = $6+4-1=9$.
• Probability = (Favorable Outcomes) / (Total Outcomes) = $9/20$.
• Wait, the provided answer is 2/5. Let’s re-check the question or my calculation.
  • Multiples of 3: 3, 6, 9, 12, 15, 18. (6)
  • Multiples of 5: 5, 10, 15, 20. (4)
  • The number 15 is in both sets.
  • The total number of unique multiples of 3 or 5 is $6+4-1=9$.
  • The probability is $9/20$. The provided answer of $2/5$ is incorrect. Let’s assume there is a typo in the question and the correct answer is $9/20$.

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7. From a pack of 52 cards, two cards are drawn together at random. What is the probability of both the cards being kings?

Answer: 1/221

• Total number of cards: 52.
• Total ways to draw 2 cards: $C(52,2)=(52\times 51)/(2\times 1)=26\times 51=1326$.
• Number of kings: 4.
• Number of ways to draw 2 kings: $C(4,2)=(4\times 3)/(2\times 1)=6$.
• Probability = (Favorable Outcomes) / (Total Outcomes) = $6/1326=\ast \ast 1/221\ast \ast$.

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8. In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither red nor green?

Answer: 1/3

• Total number of balls: $8+7+6=21$.
• Number of balls that are neither red nor green: This means the ball must be blue.
• Number of blue balls: 7.
• Probability = (Number of Blue Balls) / (Total Balls) = $7/21=\ast \ast 1/3\ast \ast$.

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9. A bag contains 2 red, 3 green and 2 blue balls. Two balls are drawn at random. What is the probability that none of the balls drawn is blue?

Answer: 10/21

• Total number of balls: $2+3+2=7$.
• Total ways to draw 2 balls: $C(7,2)=(7\times 6)/(2\times 1)=21$.
• Number of non-blue balls: $2(\text{red})+3(\text{green})=5$.
• Ways to draw 2 non-blue balls: $C(5,2)=(5\times 4)/(2\times 1)=10$.
• Probability = (Favorable Outcomes) / (Total Outcomes) = 10/21.

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10. Two dice are thrown simultaneously. What is the probability of getting two numbers whose product is even?

Answer: 3/4

• Total outcomes: $6\times 6=36$.
• For the product of two numbers to be even, at least one of the numbers must be even.
• The only case where the product is odd is when both numbers are odd.
• Let’s calculate the probability of the opposite event (product is odd) and subtract from 1.
• Odd numbers on a die are {1, 3, 5} – 3 outcomes.
• Number of ways to get an odd product: (Odd on first die) $\times$ (Odd on second die) = $3\times 3=9$.
• Probability of getting an odd product = $9/36=1/4$.
• Probability of getting an even product = $1-(\text{Probability of odd product})=1-1/4=\ast \ast 3/4\ast \ast$.