Question

Six seeds are sowed in the garden; each of which has a 70% chance of germinating.

Find the following probabilities:

a) At least two seeds will be germinated.

b) At most two seeds will be germinated

c) No seeds will be germinated

d) Mean expected number of seeds to be germinated and the corresponding St Deviation

Answer 1

Solution:

binomial distribution

n=6 p=probability of success=70%=0.7

q=1-0.7=0.3

SolutionA:

a) At least two seeds will be germinated.

P(X>=2)

=P(X=2)+P(X=3)+P(X=4)+P(X=5)+P(X=6)

=6C2*0.7^2*0.3^6-2+6C3*0.7^3*0.3^6-3+6C4*0.7^4*0.3^6-4+6C5*0.7^5*0.3^6-5+6c6*0.7^6*0.3^6-6

=0.0595+0.1852+0.3241+0.3025+0.1176

=0.9891

ANSWER:0.9891

b) At most two seeds will be germinated

P(X<=2)

=P(X=0)+P(X=1)+P(X=2)

=6C0*0.7^0*0.3^6-0+6C1*0.7^1*0.3^6-1+6C2*0.7^2*0.3^6-2

=0.000729+0.010206+0.059535

=0.07047

Solutionc:

P(X=0)

=6c0*0.7^0^0.3^6-0

=0.000729

Solutiond:

Mean expected number of seeds to be germinated=np=6*0.7=4.2

St Deviation=sqrt(npq)

=sqrt(6*0.7*0.3)

St Deviation=1.122497