[SOLVED] Binomial Probability

• Jan 28th 2010, 11:56 AM
lindah
[SOLVED] Binomial Probability
Hi,

I am having trouble with the following question:

During winter it rains on average 18 out of 30 days. Five winter days are selected at random. Find (4 d.p) the probability that:

(a) the first two days chosen will be fine and the remainder wet
My approach is this is an ordered selection:

I have defined p = it will rain = $\frac{18}{30}$ or 0.6
q = 1 - 0.6 = 0.4
= $0.4^2 \times 0.6^3$
= 0.03456

The answers I have are yielding 0.0124.

(b) more rainy days than fine days have been chosen
P(3 rainy days) = $_5C_3 \times 0.6^3 \times 0.4^2$
P(4 rainy days) = $_5C_4 \times 0.6^4 \times 0.4$
P(5 rainy days) = $_5C_5 \times 0.6^5$
= 0.3456 + 0.2592 + 0.007776
= 0.612576

I wonder where I am going wrong
TIA for any assistance!
• Jan 28th 2010, 05:01 PM
awkward
Quote:

Originally Posted by lindah
Hi,

I am having trouble with the following question:

During winter it rains on average 18 out of 30 days. Five winter days are selected at random. Find (4 d.p) the probability that:

(a) the first two days chosen will be fine and the remainder wet
My approach is this is an ordered selection:

I have defined p = it will rain = $\frac{18}{30}$ or 0.6
q = 1 - 0.6 = 0.4
= $0.4^2 \times 0.6^3$
= 0.03456 .... Looks right to me

The answers I have are yielding 0.0124.

(b) more rainy days than fine days have been chosen
P(3 rainy days) = $_5C_3 \times 0.6^3 \times 0.4^2$
P(4 rainy days) = $_5C_4 \times 0.6^4 \times 0.4$
P(5 rainy days) = $_5C_5 \times 0.6^5$
= 0.3456 + 0.2592 + 0.007776 ..... 0.07776
= 0.612576 ...... 0.68256