1. ## Binomial Distribution Problem

Hi, below is a problem under binomial distribution and i have answered the problem. Can someone advise if my solution to A below is correct please. Please help me also to solve probability for question 1B.

1. Mrs. Logan has taken 8 cuttings from a plant which won her first prize in last year’s country fair. She has set the cuttings in individual pots and hopes they will take a root. On the basis of past experience, she knows there is a 0.7 probability that a given cutting will take root.
A. Find the probability that at least 6 of the cuttings take root.
Solution:
Formula: P= nCr pr qn-r
Where:
p= 0.7
q =0.3
n = 8
r = {6,7,8}
P(X=6) = 8C6 (0.7)6 (0.3)2
= 28 (0.7)6 (0.3)2
= 28 (0.1176)(0.09)
= 0.2964
P(X=7) = 8C7 (0.7)7 (0.3)1
= 8 (0.7)7 (0.3)1
= 8 (0.0824)(0.3)
= 0.1978
P(X=8) = 8C8 (0.7)8 (0.3)0
= 1 (0.7)8 (0.3)0
= 1 (0.0576)(1)
= 0.0576
Since all events (X=6, X=7, X=8) are mutually exclusive, then
P(X>6) = P(X=6) + P(X=7) + P(X=8)
= 0.2964 + 0.1978 + 0.0576
P(X>6) = 0.5518 or 55%

B. Find the probability that more than 2 of the cuttings will take root.

For part B

$P(X > 2) = 1-P(X \leq 2) = 1 - P(X=0) - P(X=1) - P(X=2)$

Alternatively you could just lookup the answer, if you are allowed to use probability tables.

3. Originally Posted by SpringFan25

For part B

$P(X > 2) = 1-P(X \leq 2) = 1 - P(X=0) - P(X=1) - P(X=2)$

Alternatively you could just lookup the answer, if you are allowed to use probability tables.
Hi SpringFan25, thank you for your advice. Given the formula you provided for part B. Below is my calculation.

P(X=0) = 8C0 (0.7)^0 (0.3)^8
= 8 (1)(0.000066)
= 0.00053
P(X=1) = 8C1 (0.7)^1 (0.3)^7
= 8 (0.7)(0.000219)
= 0.000123
P(X=2) = 8C2 (0.7)^2 (0.3)^6
= 28 (0.49)(0.000729)
= 0.010002

P(X>2) = 1 – P(X < 2) = 1 – P(X=0) – P(X=1) – P(X=2)
= 1 – 0.00053 – 0.000123 – 0.010002
= 0.9893
P(X>2) = 1 – 0.9893 = 0.0107

Is the above computation is correct? Kindly confirm.

4. Originally Posted by jovic1104
Hi SpringFan25, thank you for your advice. Given the formula you provided for part B. Below is my calculation.

P(X=0) = 8C0 (0.7)^0 (0.3)^8
= 8 (1)(0.000066)
= 0.00053
P(X=1) = 8C1 (0.7)^1 (0.3)^7
= 8 (0.7)(0.000219)
= 0.000123
P(X=2) = 8C2 (0.7)^2 (0.3)^6
= 28 (0.49)(0.000729)
= 0.010002

P(X>2) = 1 – P(X < 2) = 1 – P(X=0) – P(X=1) – P(X=2)
= 1 – 0.00053 – 0.000123 – 0.010002
= 0.9893
ok so far (i got a diff answer but its probably rounding error)

P(X>2) = 1 – 0.9893 = 0.0107
No, you already did the 1-(stuff). You dont have to do it teice.

5. Originally Posted by SpringFan25
ok so far (i got a diff answer but its probably rounding error)

No, you already did the 1-(stuff). You dont have to do it teice.
Sorry for the error. So the final answer must be:

P(X>2) = 0.9893

I appriciate your help. Many thanks.