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?

a)P ( A∩S) =  0.116

b) P (x= 5) =0.272

c)  P (at most 5) = 0.950647

Step-by-step 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

3,750 deer

step-by-step explanation:

1,500 is 15 times larger than 100

lets make 250 15 times larger.

3750 deer