1. The probability of Anne winning a round is 2/8 or 1/4.
There are 8 possible outcomes: (H, H, H), (H, H, T), (H, T, T), (T, T, T), (T, T, H), (T, H, H), (H, T, H), and (T, H, T). Out of these 8 outcomes, Anne wins in 2 of them: (H, T, T) and (T, T, T).
Therefore, the probability of Anne winning a round is 2/8 or 1/4.
2. The probability of a round ending in a draw is 3/8.
Out of the 8 possible outcomes listed above, 3 of them result in a draw: (H, H, H), (T, T, T), and (H, T, H).
Therefore, the probability of a round ending in a draw is 3/8.
3. The probability of two draws in a row is (3/8)^2 or 9/64.
Since the outcomes of consecutive rounds are independent events, the probability of two draws in a row can be calculated by multiplying the probability of a draw in one round by itself.
Therefore, the probability of two draws in a row is (3/8)^2 or 9/64.
Three friends (Anne, Beth, and Carol) played a game which involved each of them tossing a coin. If all three coins land the same face (Heads or Tails) up, then that round ends in a draw. If one lands with a face up front different from the other two, then the one with the different face up wins the round.
Read the information above and answer the following questions below:
1. What is the probability Anne wins a round?
2. What is the probability of a round ending in a draw?
3. What is the probability of two draws in a row?
Make sure to show your work.
7 answers
Can you explain question 1. again, I don't understand your work
Of course! To find the probability of Anne winning a round, we need to consider all the possible outcomes of the game.
There are 8 total outcomes: (H, H, H), (H, H, T), (H, T, T), (T, T, T), (T, T, H), (T, H, H), (H, T, H), and (T, H, T).
Out of these 8 outcomes, Anne can win in two scenarios:
1. (H, T, T) - where Anne has a different face up than the other two, so Anne wins.
2. (T, T, T) - where Anne has the same face up as the other two, so the round ends in a draw.
Therefore, the probability of Anne winning a round is the sum of the probabilities of these two specific outcomes:
Probability of Anne winning = Probability of (H, T, T) + Probability of (T, T, T)
Probability of Anne winning = 1/8 + 1/8
Probability of Anne winning = 2/8
Probability of Anne winning = 1/4
So, the probability of Anne winning a round is 1/4 or 25%.
There are 8 total outcomes: (H, H, H), (H, H, T), (H, T, T), (T, T, T), (T, T, H), (T, H, H), (H, T, H), and (T, H, T).
Out of these 8 outcomes, Anne can win in two scenarios:
1. (H, T, T) - where Anne has a different face up than the other two, so Anne wins.
2. (T, T, T) - where Anne has the same face up as the other two, so the round ends in a draw.
Therefore, the probability of Anne winning a round is the sum of the probabilities of these two specific outcomes:
Probability of Anne winning = Probability of (H, T, T) + Probability of (T, T, T)
Probability of Anne winning = 1/8 + 1/8
Probability of Anne winning = 2/8
Probability of Anne winning = 1/4
So, the probability of Anne winning a round is 1/4 or 25%.
Can you explain question 3. again, I don't understand your work
To find the probability of two draws in a row, we need to calculate the probability of a draw in one round and then raise that probability to the power of 2 because the events of two consecutive rounds are independent.
From our previous calculations, we determined that the probability of a round ending in a draw is 3/8.
Now, to find the probability of two draws in a row, we need to square the probability of a draw in one round:
Probability of two draws in a row = Probability of a draw in one round * Probability of a draw in one round
Probability of two draws in a row = (3/8) * (3/8)
Probability of two draws in a row = 9/64
Therefore, the probability of two draws in a row is 9/64.
From our previous calculations, we determined that the probability of a round ending in a draw is 3/8.
Now, to find the probability of two draws in a row, we need to square the probability of a draw in one round:
Probability of two draws in a row = Probability of a draw in one round * Probability of a draw in one round
Probability of two draws in a row = (3/8) * (3/8)
Probability of two draws in a row = 9/64
Therefore, the probability of two draws in a row is 9/64.
How'd you get 3/8?
Apologies for the confusion earlier. Let's calculate the probability of a round ending in a draw again.
There are 8 possible outcomes: (H, H, H), (H, H, T), (H, T, T), (T, T, T), (T, T, H), (T, H, H), (H, T, H), and (T, H, T).
Out of these 8 outcomes, 3 of them result in a draw: (H, H, H), (T, T, T), and (H, T, H).
Therefore, the probability of a round ending in a draw is 3/8.
I hope this clears up the calculation for the probability of a round ending in a draw. Let me know if you have any more questions.
There are 8 possible outcomes: (H, H, H), (H, H, T), (H, T, T), (T, T, T), (T, T, H), (T, H, H), (H, T, H), and (T, H, T).
Out of these 8 outcomes, 3 of them result in a draw: (H, H, H), (T, T, T), and (H, T, H).
Therefore, the probability of a round ending in a draw is 3/8.
I hope this clears up the calculation for the probability of a round ending in a draw. Let me know if you have any more questions.
5 answers