Sorry for posting so many topics, but I really need help on this stuff. I recently lost my work I did on a separate sheet from 2 weeks ago, and now I can't seem to find the answer again.
a fair six-sided die is tossed seven times in a row.
A) What is the probability that a "4" comes up exactly four times? (I got 4375/279936)
B) What is the probability that a "4" comes up an even number of times? (I got 2315/4374)
C) What is the probability that a number higher than "4" Comes up exactly four times (I got 280/2187)
Can anyone help me with this. I need to know how I got these answers again.
Since a die is tossed seven times, for a), we want 4 comes up four times and a number other 4 three times. So,
(1/6)^4(5/6)^3
I think you can do the rest now.
not that simple
suppose we call getting a 4 is success (S) and not getting a 4 if failure (F)
one possibilty is SSSSFFF
The probability of that is (1/6)^4(5/6)^3 which is what you have.
another is SFSFSFS which would also be (1/6)^4(5/6)^3
in other words we have to count the number of arrangements of SSSSFFF which is C(7,4) or 35
So the prob of Tom's event is 35(1/6)^4(5/6)^3 or 4375/279936
which was Tom's answer.
So Tom was correct.
.40
No problem at all! I'm here to help you understand how to calculate these probabilities.
For all three questions, we need to use the concept of probability and combinatorics.
A) To find the probability that a "4" comes up exactly four times, we can break it down into two parts:
1. The probability of getting a "4" on any individual toss is 1/6.
2. The probability of not getting a "4" on any individual toss is 5/6.
We use the binomial probability formula: P(x successes) = (n choose x) * p^x * q^(n-x), where n is the number of trials, x is the number of successes, p is the probability of success, and q is the probability of failure.
In this case, n = 7 (since the die is tossed 7 times), x = 4 (since we want exactly four 4's), p = 1/6 (the probability of getting a "4"), and q = 5/6 (the probability of not getting a "4").
So the probability is P(4) = (7 choose 4) * (1/6)^4 * (5/6)^(7-4) = 35 * (1/6)^4 * (5/6)^3 = 4375/279936.
Therefore, your answer for A) is correct: P(4) = 4375/279936.
B) To find the probability that a "4" comes up an even number of times, we need to consider two cases: 0, 2, 4, or 6 "4's" appearing.
Case 0: The probability of getting no "4's" on all seven tosses is (5/6)^7.
Case 2: The probability of getting two "4's" in any two out of seven tosses is (7 choose 2) * (1/6)^2 * (5/6)^(7-2).
Case 4: The probability of getting four "4's" in any four out of seven tosses is (7 choose 4) * (1/6)^4 * (5/6)^(7-4).
Case 6: The probability of getting six "4's" in any six out of seven tosses is (7 choose 6) * (1/6)^6 * (5/6)^(7-6).
Summing up these probabilities, we get:
P(even number of 4's) = (5/6)^7 + (7 choose 2) * (1/6)^2 * (5/6)^5 + (7 choose 4) * (1/6)^4 * (5/6)^3 + (7 choose 6) * (1/6)^6 * (5/6)^1 = 2315/4374.
Therefore, your answer for B) is correct: P(even number of 4's) = 2315/4374.
C) To find the probability that a number higher than "4" comes up exactly four times, we can use the same approach as in part A), but now with a probability of 2/6 (since there are two numbers higher than 4 on a six-sided die).
So, using the binomial probability formula again, and substituting n = 7, x = 4, p = 2/6, and q = 4/6, we have:
P(highest number > 4) = (7 choose 4) * (2/6)^4 * (4/6)^(7-4) = 280/2187.
Therefore, your answer for C) is correct: P(highest number > 4) = 280/2187.
I hope this helps you understand how to calculate these probabilities. If you have any further questions, feel free to ask!