# AS maths binomial distribution question

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#1
A fair six-sided die is rolled six times. Which of the following is/are given by 6C2(1/6)^2(5/6)^4 ?
A: The probability of getting exactly one 2.
B: The probability of getting exactly two 1s.
C: The probability of getting exactly four 5s.
D: The probability of getting exactly five 4s.
E: The probability of getting exactly two scores less than 6.
F: The probability of getting exactly four scores greater than 1
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#2
(Original post by jmyjmy)
A fair six-sided die is rolled six times. Which of the following is/are given by 6C2(1/6)^2(5/6)^4 ?
A: The probability of getting exactly one 2.
B: The probability of getting exactly two 1s.
C: The probability of getting exactly four 5s.
D: The probability of getting exactly five 4s.
E: The probability of getting exactly two scores less than 6.
F: The probability of getting exactly four scores greater than 1
0
4 months ago
#3
(Original post by jmyjmy)
A fair six-sided die is rolled six times. Which of the following is/are given by 6C2(1/6)^2(5/6)^4 ?
A: The probability of getting exactly one 2.
B: The probability of getting exactly two 1s.
C: The probability of getting exactly four 5s.
D: The probability of getting exactly five 4s.
E: The probability of getting exactly two scores less than 6.
F: The probability of getting exactly four scores greater than 1
Its obviously from a binomial distribution, so any ideas?
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#4
(Original post by mqb2766)
Its obviously from a binomial distribution, so any ideas?
I can see that B would be 6C2(1/6)^2(5/6)^4, but I'm confused about E and F
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4 months ago
#5
(Original post by jmyjmy)
I can see that B would be 6C2(1/6)^2(5/6)^4, but I'm confused about E and F
Agree about that. For the last two, how would you write down the probabilities? Are they equivalent to any of A-D?
Last edited by mqb2766; 4 months ago
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#6
(Original post by mqb2766)
Agree about that. For the last two, how would you write down the probabilities?
For E, would it be 6C2(5/6)^2(1/6)^4
And for F would it be 6C4(5/6)^4(1/6)^2
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4 months ago
#7
(Original post by jmyjmy)
For E, would it be 6C2(5/6)^2(1/6)^4
And for F would it be 6C4(5/6)^4(1/6)^2
And you know 6C2 = 6C4?
F must be the same as B, even if you don't understand probabilities?
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#8
(Original post by mqb2766)
Agree about that. For the last two, how would you write down the probabilities? Are they equivalent to any of A-D?
I don't think the last two would be equivalent to any of A-D, because the success probability is 1/6 for A to D, but this changes to 5/6 for E and F
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#9
(Original post by mqb2766)
And you know 6C2 = 6C4?
F must be the same as B, even if you don't understand probabilities?
OHHH I see.. thank you so much I understand now
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4 months ago
#10
(Original post by jmyjmy)
I don't think the last two would be equivalent to any of A-D, because the success probability is 1/6 for A to D, but this changes to 5/6 for E and F
Can you see which one E is equivalent to? It's not identical, but essentially the same.
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#11
(Original post by mqb2766)
Can you see which one E is equivalent to? It's not identical, but essentially the same.
Yes, is the probability of E equivalent to C?
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4 months ago
#12
(Original post by jmyjmy)
Yes, is the probability of E equivalent to C?
Yes. When using a binomial, you can go down the p or 1-p route, and note the nCr symmetry
nCr = nC(n-r)
Which is the symmetry in Pascal triangle.
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#13
(Original post by mqb2766)
Yes. When using a binomial, you can go down the p or 1-p route, and note the nCr symmetry
nCr = nC(n-r)
Which is the symmetry in Pascal triangle.
Since nCr = nC(n-r) is true, 6C2 = 6C4 = 15
Also, the row in Pascal's triangle when n = 6 is 1, 6, 15, 20, 15, 6, 1
Therefore it's 15 for the 2nd and 4th term in this row of Pascal's triangle!
Thanks so much for the help - I'm starting to love the statistics side of maths
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