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Hi! I have a question. I was solving 2 very similar problems, where one needed to account for premutations, while other didn't. Can someone help explain why?
Problem 1 => Accounted for permutations
Q: 4 coin flips. Probability of landing at least 3 heads?
AC:
Listing all possible cases:
HHHH => (1/2)^4
HHHT => (1/2)^4 * (4!/3!*1!)
HHTT => invalid
HTTT => invalid
TTTT => invalid
Problem 2 => Did not need to account for permutations
Q: Going to buy 8 bags of rice in the store, each from Brand A or Brand B. Probability of buying at least 2 bags of A or at least 2 bags of B?
AC:
Listing all possible cases:
AAAAAAAA => Invalid
AAAAAAAB => Invalid
AAAAAABB => Valid case
AAAAABBB => Valid case
AAAABBBB => Valid case
AAABBBBB => Valid case
AABBBBBB => Valid case
ABBBBBBB => Valid case
BBBBBBBB => Invalid
5 valid cases out of 9 total cases => probability of 5/9
Link to quetsions
https://gmatclub.com/forum/a-fair-coin- ... 99478.html
https://gmatclub.com/forum/a-customer-w ... 64314.html
Problem 1 => Accounted for permutations
Q: 4 coin flips. Probability of landing at least 3 heads?
AC:
Listing all possible cases:
HHHH => (1/2)^4
HHHT => (1/2)^4 * (4!/3!*1!)
HHTT => invalid
HTTT => invalid
TTTT => invalid
Problem 2 => Did not need to account for permutations
Q: Going to buy 8 bags of rice in the store, each from Brand A or Brand B. Probability of buying at least 2 bags of A or at least 2 bags of B?
AC:
Listing all possible cases:
AAAAAAAA => Invalid
AAAAAAAB => Invalid
AAAAAABB => Valid case
AAAAABBB => Valid case
AAAABBBB => Valid case
AAABBBBB => Valid case
AABBBBBB => Valid case
ABBBBBBB => Valid case
BBBBBBBB => Invalid
5 valid cases out of 9 total cases => probability of 5/9
Link to quetsions
https://gmatclub.com/forum/a-fair-coin- ... 99478.html
https://gmatclub.com/forum/a-customer-w ... 64314.html
17 Oct 2024, 13:45
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GMATkid1997
I think you're missing a key point in both of these questions.
Question 1: A fair coin is tossed 5 times. What is the probability of getting at least three heads on consecutive tosses?
In this question, notice that it specifically asks for "at least three heads on consecutive tosses." So, we're focusing on consecutive heads, not just any combination.
For at least 3 heads, we need to consider cases with 3, 4, or 5 consecutive heads.
3 consecutive heads: There are 5 possible cases:
- HHHTT
- THHHT
- TTHHH
- HTHHH
- HHHTH
The probability for each is (1/2)^5, so: P = 5 * (1/2)^5 = 5/32.
4 consecutive heads: There are 2 possible cases:
- HHHHT
- THHHH
The probability for each is (1/2)^5, so: P = 2 * (1/2)^5 = 2/32.
5 consecutive heads: There is 1 possible case:
- HHHHH
The probability is (1/2)^5, so: P = 1/32.
Adding them up: P = 5/32 + 2/32 + 1/32 = 8/32 = 1/4.
Question 2: A customer wants to buy a total of 8kg rice of two varieties- A and B. If he can buy each type of rice only in integral multiples of 1 kg then what is the probability that he will buy at least 2 kg rice of each variety?
This question is not about the arrangement at all! The focus is on how to divide the 8 kg between two varieties, A and B, ensuring that at least 2 kg of each is bought.
There are 9 total ways to divide 8 kg between A and B:
- 0 - 8
- 1 - 7
- 2 - 6
- 3 - 5
- 4 - 4
- 5 - 3
- 6 - 2
- 7 - 1
- 8 - 0
Out of these, only 5 combinations meet the requirement of having at least 2 kg of each variety. So, the probability is 5/9.
