Math Problem Statement

A university delivers a niche degree programme, aiming to recruit a small number of students each year.

Applications to the programme are received at an average rate of 4.5 per year.

When the university makes offers to applicants, each offer holder accepts their offer with probability 0.65 .

When answering this question you may assume that applications are received by the department independently, at a constant average rate. You may also assume that each of the offer holders makes the decision whether or not to accept their offer independently of all other offer holders.

Questions a. Calculate the probability that the degree programme receives at least four applications next year.

You should give your answer correct to at least four significant figures. This part of the question is worth 1 mark.

b. In one year the programme makes six offers.

Calculate the probability that at least four offer holders accept their offer.

You should give your answer correct to at least two significant figures. This part of the question is worth 2 marks.

Answers

Solution

Part (a): Probability that the degree programme receives at least 4 applications next year

Given:

  • The average rate of applications is 4.5 per year.
  • The number of applications follows a Poisson distribution with mean λ=4.5\lambda = 4.5.

To calculate the probability that the programme receives at least 4 applications, we need to find:

P(X4)P(X \geq 4)

Where XX follows a Poisson distribution with parameter λ=4.5\lambda = 4.5. The formula for the Poisson probability mass function is:

P(X=k)=λkeλk!P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}

We can calculate P(X4)P(X \geq 4) as:

P(X4)=1P(X<4)=1(P(X=0)+P(X=1)+P(X=2)+P(X=3))P(X \geq 4) = 1 - P(X < 4) = 1 - (P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3))

We'll now calculate the individual probabilities.

Step-by-step calculations:

P(X=0)=4.50e4.50!=e4.5P(X = 0) = \frac{4.5^0 e^{-4.5}}{0!} = e^{-4.5} P(X=1)=4.51e4.51!=4.5e4.5P(X = 1) = \frac{4.5^1 e^{-4.5}}{1!} = 4.5 e^{-4.5} P(X=2)=4.52e4.52!=4.52e4.52P(X = 2) = \frac{4.5^2 e^{-4.5}}{2!} = \frac{4.5^2 e^{-4.5}}{2} P(X=3)=4.53e4.53!=4.53e4.56P(X = 3) = \frac{4.5^3 e^{-4.5}}{3!} = \frac{4.5^3 e^{-4.5}}{6}

Using the exponential function e4.50.0111089965e^{-4.5} \approx 0.0111089965, we compute these values.

Final result:

After computing these probabilities, we subtract the sum from 1 to get the final probability P(X4)P(X \geq 4).

Part (b): Probability that at least four offer holders accept their offer

Given:

  • The programme makes 6 offers.
  • Each offer holder accepts their offer with probability 0.65.
  • The number of acceptances follows a Binomial distribution, where the number of trials is n=6n = 6 and the probability of success (accepting the offer) is p=0.65p = 0.65.

We need to calculate the probability that at least 4 offer holders accept their offer, i.e., P(Y4)P(Y \geq 4), where YBinomial(n=6,p=0.65)Y \sim \text{Binomial}(n = 6, p = 0.65).

The formula for the Binomial distribution is:

P(Y=k)=(nk)pk(1p)nkP(Y = k) = \binom{n}{k} p^k (1 - p)^{n-k}

We want:

P(Y4)=1P(Y3)=1(P(Y=0)+P(Y=1)+P(Y=2)+P(Y=3))P(Y \geq 4) = 1 - P(Y \leq 3) = 1 - (P(Y = 0) + P(Y = 1) + P(Y = 2) + P(Y = 3))

We'll compute each of these binomial probabilities:

P(Y=k)=(6k)(0.65)k(0.35)6kP(Y = k) = \binom{6}{k} (0.65)^k (0.35)^{6-k}

After computing this, we'll subtract the sum from 1 to get the final probability.


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Math Problem Analysis

Mathematical Concepts

Poisson Distribution
Binomial Distribution
Probability Theory

Formulas

Poisson distribution probability mass function: P(X = k) = (λ^k * e^(-λ)) / k!
Binomial distribution probability mass function: P(Y = k) = (n choose k) * p^k * (1 - p)^(n - k)

Theorems

Poisson distribution
Binomial distribution

Suitable Grade Level

Grades 11-12