Practice Problems (Junior Cert Mathematics): Revision Notes

Practice Problems

Problems:

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Problem 1: Flipping Coins

Explanation: When flipping coins, each flip is independent, meaning what happens on one flip doesn't affect the other. To find the probability for multiple flips, we think about the outcomes of each coin and combine them.

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Problem 2: Rolling a Die

Explanation: Rolling a die has 6 possible outcomes. When you roll it twice, each roll is independent, so the outcomes combine just like with flipping coins.

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Problem 3: Picking Marbles

Explanation:

When you pick a marble and put it back, the chances stay the same each time because the total number of marbles doesn't change. This is called "with replacement."

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Problem 4: Rolling Dice and Flipping a Coin

Explanation: When you combine two different actions, like rolling a die and flipping a coin, you think about the possible outcomes of each action and then combine them.

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Problem 5: Drawing Cards

Explanation:

When you draw a card and then replace it back into the deck before drawing again, the total number of cards remains the same each time. This means the probability for each draw stays consistent.

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Solutions

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Problem 1: Flipping Coins

1. You flip two coins. What is the probability of getting a head on both coins?

• Solution:
• Step 1: The chance of getting a head on one coin is $\frac{1}{2}$. This is because there are two possible outcomes when you flip a coin: heads or tails, and only one of those is heads.
• Step 2: Since you want heads on both coins, and each flip is separate (independent), you use the "AND" rule. This means you multiply the probability of getting heads on the first coin by the probability of getting heads on the second coin: $P(\text{Head AND Head}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$
• Explanation: We multiply because each flip of the coin does not affect the other. This is why the probability of getting heads on both coins is smaller than just one flip.
• Answer: The chance of getting heads on both coins is $\frac{1}{4}$, or 25%.

2. You flip two coins. What is the probability of getting a head on at least one of the coins?

• Solution:
• Step 1: First, let's think about the opposite outcome, which is getting tails on both coins. The probability of getting tails on one coin is $\frac{1}{2}$.
• Step 2: Since both flips are independent, you use the "AND" rule to multiply the probabilities: $P(\text{Tails AND Tails}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$
• Step 3: Now, since you want the opposite (at least one head), subtract this result from 1. This works because the probability of all possible outcomes must add up to 1: $P(\text{At least one head}) = 1 - \frac{1}{4} = \frac{3}{4}$
• Explanation: Subtracting from 1 helps us find the chance of getting at least one head by eliminating the only case where no heads are flipped.
• Answer: The chance of getting a head on at least one coin is $\frac{3}{4}$, or 75%.

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Problem 2: Rolling a Die

3. You roll a fair six-sided die twice. What is the probability of rolling a $4$ on both rolls?

• Solution:
• Step 1: The chance of rolling a $4$ on one roll of a die is $\frac{1}{6}$. This is because a die has six sides, and only one of those sides is a $4$.
• Step 2: Since you want to roll a $4$ on both rolls, and each roll is separate (independent), use the "AND" rule to multiply the probabilities: $P(\text{4 AND 4}) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$
• Explanation: Just like with coins, each roll is independent, so you multiply the probabilities to find the chance of both events happening.
• Answer: The chance of rolling a 4 on both rolls is $\frac{1}{36}$, or about 2.78%.

4. You roll a fair six-sided die twice. What is the probability of rolling a $4$ on the first roll or a $5$on the second roll?

• Solution:
• Step 1: The chance of rolling a $4$ on the first roll is $\frac{1}{6}$.
• Step 2: The chance of rolling a $5$ on the second roll is also $\frac{1}{6}$.
• Step 3: Since either one of these can happen (you want one or the other), use the "OR" rule and add the probabilities: $P(\text{4 on 1st OR 5 on 2nd}) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$
• Explanation: Adding the probabilities works here because you're looking for either outcome, not both together.
• Answer: The chance of rolling a $4$on the first roll or a $5$ on the second roll is $\frac{1}{3}$, or about 33.33%.

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Problem 3: Picking Marbles

5. You have a bag containing $3$ $red$ $marbles$ and $5$ $blue$ $marbles$. If you pick one marble, put it back, and then pick another marble, what is the probability of picking $two$ $red$ $marbles$ in a row?

• Solution:
• Step 1: The chance of picking a red marble on the first draw is $\frac{3}{8}$. This is because there are $3$ $red$ $marbles$ out of $8$ $total$ $marbles$.
• Step 2: After putting the marble back, the chance of picking a red marble again stays the same at $\frac{3}{8}$.
• Step 3: Since you want both events to happen (picking a red marble twice), use the "AND" rule to multiply the probabilities: $P(\text{Red AND Red}) = \frac{3}{8} \times \frac{3}{8} = \frac{9}{64}$
• Explanation: We multiply because putting the marble back makes each pick independent, just like flipping a coin.
• Answer: The chance of picking $two$ $red$ $marbles$ in a row is $\frac{9}{64}$, or about 14.06%.

6. You have a bag containing $3$ $red$ $marbles$ and $5$ $blue$ $marbles$. If you pick one marble, put it back, and then pick another marble, what is the probability of picking a $red$ $marble$ first, followed by a $blue$ $marble$?

• Solution:
• Step 1: The chance of picking a red marble first is $\frac{3}{8}$.
• Step 2: The chance of picking a blue marble second is $\frac{5}{8}$.
• Step 3: Since you want both events to happen in this order ($red$ first, $blue$ second), use the "AND" rule to multiply the probabilities: $P(\text{Red AND Blue}) = \frac{3}{8} \times \frac{5}{8} = \frac{15}{64}$
• Explanation: Multiplying helps us find the combined probability of two independent events happening one after the other.
• Answer: The chance of picking a red marble first and a blue marble second is $\frac{15}{64}$, or about 23.44%.

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Problem 4: Rolling

Dice and Flipping a Coin**

7. You roll a six-sided die and flip a coin. What is the probability of rolling an even number and getting a head?

• Solution:
• Step 1: The even numbers on a die are $2, 4,$ and $6$. So the chance of rolling an even number is $\frac{3}{6} = \frac{1}{2}$.
• Step 2: The chance of getting a head on the coin is $\frac{1}{2}$.
• Step 3: Since you want both things to happen (rolling an even number and getting a head), use the "AND" rule to multiply the probabilities: $P(\text{Even AND Head}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$
• Explanation: Multiplying tells us the chance of both independent events happening together.
• Answer: The chance of rolling an even number and getting a head is $\frac{1}{4}$, or 25%.

8. You roll a six-sided die and flip a coin. What is the probability of rolling a number less than $3$ or getting a tail?

• Solution:
• Step 1: The numbers less than $3$ on a die are $1$ and $2$. So the chance of rolling a number less than $3$ is $\frac{2}{6} = \frac{1}{3}$.
• Step 2: The chance of getting a tail on the coin is $\frac{1}{2}$.
• Step 3: Since you want either one of these outcomes to happen, use the "OR" rule and add the probabilities: $P(\text{Less than 3 OR Tail}) = \frac{1}{3} + \frac{1}{2} = \frac{2}{6} + \frac{3}{6} = \frac{5}{6}$
• Explanation: Adding the probabilities gives us the total chance of either event happening.
• Answer: The chance of rolling a number less than $3$ or getting a tail is $\frac{5}{6}$, or about 83.33%.

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Problem 5: Drawing Cards

9. A deck of cards has $52$ $cards$ ($13$ of each suit: $hearts$, $diamonds$, $clubs$, and $spades$). If you draw one card, replace it, and then draw another card, what is the probability of drawing $two$ $aces$ in a row?

• Solution:
• Step 1: The chance of drawing an ace on the first draw is $\frac{4}{52} = \frac{1}{13}$. This is because there are $4$ $aces$ out of $52$ $cards$.
• Step 2: After replacing the card, the chance of drawing an ace again remains $\frac{1}{13}$.
• Step 3: Since you want both events to happen ($drawing$ $two$ $aces$), use the "AND" rule to multiply the probabilities: $P(\text{Ace AND Ace}) = \frac{1}{13} \times \frac{1}{13} = \frac{1}{169}$
• Explanation: Multiplying finds the chance of drawing an ace both times when the card is replaced.
• Answer: The chance of drawing $two$ $aces$ in a row is $\frac{1}{169}$, or about 0.59%.

10. A deck of cards has $52$ $cards$ ($13$ of each suit: $hearts$, $diamonds$, $clubs$, and $spades$). If you draw one card, replace it, and then draw another card, what is the probability of drawing a $red$ $card$ ($hearts$ or $diamonds$) followed by a black card ($clubs$ or $spades$)?

• Solution:
• Step 1: The chance of drawing a $red$ $card$ ($hearts$ or $diamonds$) on the first draw is $\frac{26}{52} = \frac{1}{2}$.
• Step 2: The chance of drawing a $black$ $card$ ($clubs$ or $spades$) on the second draw is also $\frac{1}{2}$.
• Step 3: Since you want both events to happen ($drawing$ $a$ $red$ $card$ $first$ and $a$ $black$ $card$ $second$), use the "AND" rule to multiply the probabilities: $P(\text{Red AND Black}) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$
• Explanation: Multiplying tells us the chance of drawing one colour first, then the other, when the card is replaced.
• Answer: The chance of drawing $a$ $red$ $card$ followed by $a$ $black$ $card$ is $\frac{1}{4}$, or 25%.