In an experiment with x-ray radiation of microspores a study(Catcheside et al.,

Question

In an experiment with x-ray radiation of microspores a study(Catcheside et al., 1945) found

that the number of chromosomal changes in cells in a certain typeof experiment follows a

Poisson distribution with mean 0.480.

a. Compute the probability that zero changes are observed in suchan experiment.

b. Find the probability that greater than 3 changes areobserved.

c. In a particular experiment, suppose it is known that the numberof chromosomal

changes in cells is larger than three. Find the probabilitythat five chromosomal

changes in cells are observed.

d. Suppose 10 such experiments are conducted and the total numberof chromosomal

changes in cells is computed. What is the probability ofobserving a total of 3 or

more chromosomal changes in cells?

Explanation / Answer

a)      For Poisson distribution with mean= 0.48

P(x=k) = (k * e-)/k!

Therefore for k =0 we get P(we have zero changes in the time) =e-0.48 = 61.87%

b)      P(greater than 3 changes areobserved) = 1- P(lesser or equal to 3 changes are observed)

Advantage of doing it this way is we can get away by calculating4 terms.

= 1- P(x=0)-P(x=1)-P(x=2)-P(x=3)

=1-(0 * e-)/0!-(1 * e-)/1!-(k2* e-)/2!-(3 * e-)/3!

= 1-99.848% = 0.1511%

c)       Let us define two events

A: observing more than 3 changes

B: observing 5 changes

P(B|A) = P(A &B)/P(A)

= P(X=5)/P(x>3)

=(5 * e-)/5!/0.1511%

= 0.0131%

=8.68%

d)      P(total changes observed inall 10 experiments> =3 ) =1-P(total changes observed<3)

We have to calculate the probabilities of total =0 ,1,2 and addthem to get the required probability

P(Total X =0) : only one way to get 0 i.e. when we observex=0 in all experiments = 61.87%10 = 0.823%

P(total X= 1) : we can choose that one experiment thatgives 1 change in 10c1 = 10 ways

And probability = P(x=1)*(P(x=0))9 =61.78%9 *29.70% =0.392%

P(total x =2): we can get this in two different ways. Fist whentwo experiments show 1 change each and all the other 8 show 0changes. Second when one experiment show 2 changes and rest 9 show0 changes.

P(total x=2) = 10c2* P(x=1)2*P(x=0)8 +10c1 * P(x=2)*P(x=0)9

=45*29.70%2 *61.87%8 +10*7.128%*61.87%9 = 9.48%

Therefore required probability = 1- 0.823%-0.392%-9.48% =89.30%

Hope this helps you. Feel free to ask for any clarification

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.