QuestionWhat is the probability of rolling greater than 1 on a die and getting heads on a coin? Answer as a fraction.

Studdy Solution

STEP 1

Assumptions1. The die is fair, which means each of its six faces (1,,3,4,5,6) has an equal probability of1/6. . The coin is fair, which means it has two outcomes (heads, tails) each with an equal probability of1/.
3. The events of rolling the die and flipping the coin are independent, meaning the outcome of one does not affect the outcome of the other.

STEP 2

First, we calculate the probability of rolling a number greater than1 on the die. This includes the numbers2,,4,5, and6. Since each number has an equal probability of1/6, the probability of rolling a number greater than1 is the sum of the probabilities of each of these outcomes.
$(Die >1) =(2) +() +(4) +(5) +(6)$

STEP 3

Substitute the probability of each outcome (1/6) into the equation.
$(Die >1) =5 \times \frac{1}{6}$

STEP 4

Calculate the probability of rolling a number greater than1.
$(Die >1) = \frac{}{6}$

STEP 5

Next, we calculate the probability of getting heads on the coin. Since the coin is fair, this is simply1/2.
$(Heads) = \frac{1}{2}$

STEP 6

Since the events are independent, the probability of both events happening is the product of their individual probabilities. $(Die >1 \, and \, Heads) =(Die >1) \times(Heads)$

STEP 7

Substitute the calculated probabilities into the equation.
$(Die >1 \, and \, Heads) = \frac{5}{6} \times \frac{1}{2}$

STEP 8

Calculate the probability of both events happening.
$(Die >1 \, and \, Heads) = \frac{5}{12}$ The probability that you roll a number greater than1 and get heads with the coin is5/12.

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QuestionWhat is the probability of rolling greater than 1 on a die and getting heads on a coin? Answer as a fraction.

Studdy Solution

STEP 1

STEP 2

STEP 3

Substitute the probability of each outcome (1/6) into the equation.
$(Die >1) =5 \times \frac{1}{6}$

STEP 4

Calculate the probability of rolling a number greater than1.
$(Die >1) = \frac{}{6}$

STEP 5

Next, we calculate the probability of getting heads on the coin. Since the coin is fair, this is simply1/2.
$(Heads) = \frac{1}{2}$

STEP 6

Since the events are independent, the probability of both events happening is the product of their individual probabilities. $(Die >1 \, and \, Heads) =(Die >1) \times(Heads)$

STEP 7

Substitute the calculated probabilities into the equation.
$(Die >1 \, and \, Heads) = \frac{5}{6} \times \frac{1}{2}$

STEP 8

Calculate the probability of both events happening.
$(Die >1 \, and \, Heads) = \frac{5}{12}$ The probability that you roll a number greater than1 and get heads with the coin is5/12.

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QuestionWhat is the probability of rolling greater than 1 on a die and getting heads on a coin? Answer as a fraction.

Studdy Solution

STEP 1

STEP 2

STEP 3

Substitute the probability of each outcome (1/6) into the equation.(Die>1)=5×16(Die >1) =5 \times \frac{1}{6}(Die>1)=5×61​

STEP 4

Calculate the probability of rolling a number greater than1.(Die>1)=6(Die >1) = \frac{}{6}(Die>1)=6​

STEP 5

Next, we calculate the probability of getting heads on the coin. Since the coin is fair, this is simply1/2.(Heads)=12(Heads) = \frac{1}{2}(Heads)=21​

STEP 6

Since the events are independent, the probability of both events happening is the product of their individual probabilities.(Die>1 and Heads)=(Die>1)×(Heads)(Die >1 \, and \, Heads) =(Die >1) \times(Heads)(Die>1andHeads)=(Die>1)×(Heads)

STEP 7

Substitute the calculated probabilities into the equation.(Die>1 and Heads)=56×12(Die >1 \, and \, Heads) = \frac{5}{6} \times \frac{1}{2}(Die>1andHeads)=65​×21​

STEP 8

Calculate the probability of both events happening.(Die>1 and Heads)=512(Die >1 \, and \, Heads) = \frac{5}{12}(Die>1andHeads)=125​The probability that you roll a number greater than1 and get heads with the coin is5/12.

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Substitute the probability of each outcome (1/6) into the equation.
$(Die >1) =5 \times \frac{1}{6}$

Calculate the probability of rolling a number greater than1.
$(Die >1) = \frac{}{6}$

Next, we calculate the probability of getting heads on the coin. Since the coin is fair, this is simply1/2.
$(Heads) = \frac{1}{2}$

Since the events are independent, the probability of both events happening is the product of their individual probabilities. $(Die >1 \, and \, Heads) =(Die >1) \times(Heads)$

Substitute the calculated probabilities into the equation.
$(Die >1 \, and \, Heads) = \frac{5}{6} \times \frac{1}{2}$

Calculate the probability of both events happening.
$(Die >1 \, and \, Heads) = \frac{5}{12}$ The probability that you roll a number greater than1 and get heads with the coin is5/12.