Question

1. The number of missed notes by a musician in a symphony is Poisson. Suppose the principle trombonist misses an average of 2 notes per concert, and the 2nd trombonist misses an average of 5 notes per concert.

(a) What is the probability that they miss no notes in a concert?

(b) Suppose a concert requires only one trombonist and there is an even chance either is chosen. What is the probability no notes are missed in that concert?

(c) If no notes were missed, what is the probability that the principle trombonist was playing?

Answer 1

(a)

Let X and Y be the number of missed notes per concert by principle trombonist and 2nd trombonist respectively.

X ~ Poisson(2)

Y ~ Poisson(5)

Probability that they principle trombonist miss no notes in a concert = P(X = 0) = = 0.1353

Probability that they 2nd trombonist miss no notes in a concert = P(Y = 0) = = 0.0067

Probability that they miss no notes in a concert = P(X = 0) * P(Y = 0) (X and Y are independent)

= 0.1353 * 0.0067

= 0.0009

(b)

Let N be the event that no notes are missed in that concert.

Let PT and 2T be the event that principle trombonist and 2nd trombonist respectively are chosen for the concert.

P(PT) = P(2T) = 0.5

P(N | PT) = 0.1353

P(N | 2T) = 0.0067

By law of total probability,

P(N) = P(N | PT) P(PT) + P(N | 2T) P(2T)

= 0.1353 * 0.5 + 0.0067 * 0.5

= 0.071

(c)

If no notes were missed, probability that the principle trombonist was playing = P(PT | N)

= P(N | PT) P(PT) / P(PT) (Bayes Theorem)

= 0.1353 * 0.5 / 0.071

= 0.9528