your answer to part A is correct
For part B
Alternatively you could just lookup the answer, if you are allowed to use probability tables.
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.
A. Find the probability that at least 6 of the cuttings take root.
- 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.
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.
Thanks in advance.
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.