you flip a coin and toss a 1-6 number cube.
P(not tails and not a 3
P (Not tails) = P (heads)
P (not 3) = P (1 + 2 + 4 + 5 + 6) = 1/6 + 1/6....
Probability of all/both events occurring = product of individual events.
1/12
To find the probability of not getting tails and not getting a 3 when flipping a coin and tossing a 1-6 number cube, we need to determine the probabilities of each event separately and then multiply them together.
Let's break down the problem:
1. Flipping a coin: When flipping a fair coin, there are two possible outcomes - heads (H) or tails (T). The probability of getting tails is 1/2, and the probability of not getting tails (i.e., getting heads) is also 1/2.
2. Tossing a 1-6 number cube: When tossing a fair 6-sided die, there are six possible outcomes - numbers 1, 2, 3, 4, 5, or 6. The probability of getting a 3 is 1/6, and therefore, the probability of not getting a 3 is 1 - 1/6 = 5/6.
To find the probability of not getting tails and not getting a 3, we multiply the individual probabilities together:
P(not tails and not a 3) = P(not tails) × P(not a 3)
= (1/2) × (5/6)
= 5/12
Therefore, the probability of not getting tails and not getting a 3 is 5/12.