Marcus flips a coin and tosses a six-sided die with the sides numbered 1-6. What is the probaility that he gets a head on the coin and a number divisible by 3 on the die?
A six-sided die with sides numbered 1 through 6 is tossed twice. What is the probability of getting a number larger than 4 on both throws?
1/9
Elizabeth:
3 and 6 are divisible by 3 so 2/6 = 1/3
1/2 *1/3 = 1/6
Anon:
5 or 6/ 6
1/3 each time
independent tosses so
1/3 * 1/3 = 1/9
To find the probability that Marcus gets a head on the coin and a number divisible by 3 on the die, we need to calculate the number of favorable outcomes and divide it by the total number of possible outcomes.
Let's start by determining the number of favorable outcomes.
For the coin toss, there are two possible outcomes: heads (H) or tails (T). Since we are interested only in heads, we have one favorable outcome for the coin toss.
Next, let's consider the die roll. Out of the six sides, there are two numbers that are divisible by 3, which are 3 and 6. Therefore, we have two favorable outcomes for the die roll.
To find the total number of possible outcomes, we multiply the number of outcomes for each event. In this case, since the coin toss and die roll are independent events, we can simply multiply the number of outcomes for each event.
The coin toss has 2 possible outcomes, and the die roll has 6 possible outcomes. Thus, the total number of possible outcomes is 2 (coin toss) multiplied by 6 (die roll), which equals 12.
So, we have 1 favorable outcome and 12 possible outcomes.
To find the probability, we divide the number of favorable outcomes (1) by the number of possible outcomes (12):
Probability = Number of favorable outcomes / Number of possible outcomes
= 1 / 12
= 1/12
Therefore, the probability that Marcus gets a head on the coin and a number divisible by 3 on the die is 1/12.