Question

Forty percent of seeds from maize (modern-day corn) ears carry single spikelets, and the other 60% carry paired spikelets. A seed with single spikelets will produce an ear with single spikelets 29% of the time, whereas a seed with paired spikelets will produce an ear with single spikelets 26% of the time. Consider randomly selecting ten seeds.

a. What is the probability that exactly five of these seeds carry a single spikelet and produce an ear with a single spikelet?

b. What is the probability that exactly five of the ears produced by these seeds have single spikelets? What is the probability that at most five ears have single spikelets?

Answer 1

a)P ( A∩S) = 0.116

b) P (x= 5) =0.272

c) P (at most 5) = 0.950647

Explanation:

Let A be an event that the maize (modern-day corn) ears carry single spikelets and B be an event that the maize carry paired spikelets.

Then P (B) = 60% = 0.6

P (A) = 1-P (b) = 0.4

Let S be the event that seed with single spikelets will produce an ear with single spikelets and Q be the event that seed with paired spikelets will produce an ear with single spikelets .

Then

P (S) = 29% = 0.29

P (Q ) = 26% = 0.26

n= 10

We want to find out that exactly 5 single spikelets seeds produce single spikelets .

P ( A∩S) = P(A) *P(S) = 0.4 * 0.29= 0.116

Now we have to find that exactly five of the ears produced by these seeds have single spikelets.

b) P (x= 5) = P(A) *P(S)+ P(B) *P(R)

= 0.4 * 0.29 + 0.6 * 0.26

= 0.116+ 0.156= 0.272

At most five ears have single spikelets.This follows a binomial distribution where n is fixed and also p and q are same.

P ( x= at most 5) = 10C5 ( 0.116)^ 5(0.884)^5 +10C4 ( 0.116)^ 4(0.884)^6+10C3 ( 0.116)^ 3(0.884)^7+10C2 ( 0.116)^ 2(0.884)^8 +10C9 ( 0.116)^ 1(0.884)^9+10C10 ( 0.116)^ 0(0.884)^10

=0.002857+0.018145+ 0.0000234+0.2558+ 0.3824+ 0.29144

= 0.950647