Answer to Question #310398 in Statistics and Probability for Subham

Question

From a population of 700 plants, 500 plants are healthy, rest are unhealthy. For a sample size of 10 plants, calculate the probabilities that—(a) Exactly 3 are unhealthy (b) at least 3 plants are unhealthy (c) at most 3 plants are unhealthy (d) Exactly 3 are healthy (e) None healthy?

Answer 1

Let X be the random variable representing the number of unhealthy plants.

The probability that a plant is unhealthy p=200/700 = 2/7

Since each plant can either be healthy or unhealthy, there are only two outcomes. Therefore, X follows a binomial distribution given by the formula:

P(X=x) =_nC_xp^x(1-p)^n-x

For x = 0,1,2...,n.

For n=10 and p=2/7=0.2857

a)

P(X=3) = ₁₀C₃(0.2857)³(0.7143)⁷

=0.2655

b)

P(X≥3) = P(3)+ P(4)+P(5)+P(6)+P(7)+P(8)+P(9)+P(10)

P(X=3) =₁₀C₃(0.2857)³(0.7143)⁷=0.2655

P(X=4) =₁₀C₄(0.2857)⁴(0.7143)⁶=0.1858

P(X=5) =₁₀C₅(0.2857)⁵(0.7143)⁵=0.0892

P(X=6) =₁₀C₆(0.2857)⁶(0.7143)⁴=0.0297

P(X=7)=₁₀C₇(0.2857)⁷(0.7143)³= 0.0068

P(X=8) =₁₀C₈(0.2857)⁸(0.7143)²= 0.0010

P(X=9) =₁₀C₉(0.2857)⁹(0.7143)¹=0

P(X=0) =₁₀C₁₀(0.2857)¹⁰(0.7143)⁰=0

Therefore;

P(X≥3)=0.2655+0.1858+0.0892+0.0297+0.0068+0.0010

=0.5782

c)

P(X ≤ 3)= P(0)+P(1)+P(2)+P(3)

P(X=0)=₁₀C₀(0.2857)⁰(0.7143)¹⁰=0.0346

P(X=1)=₁₀C₁(0.2857)¹(0.7143)⁹= 0.1383

P(X=2) =₁₀C₂(0.2857)²(0.7143)⁸= 0.2489

P(X=3) =₁₀C₃(0.2857)³(0.7143)⁷=0.2655

Therefore,

P(X≤3)=0.0346+0.1383+0.2489+0.2655

=0.6873

d)

Let the probability of a healthy plant p=5/7=0.7143

Let U be the random variable representing the number of healthy plants.

Then;

P(U=3) =₁₀C₃(0.7143)³(0.2857)⁷

=0.0068

e)

P(none is healthy) =P(all are healthy) P(X=10) =₁₀C₁₀(0.2857)¹⁰(0.7143)⁰

=0