Homepage › Solution manuals › Joseph Blitzstein › Introduction to Probability › Exercise 3.33

Exercise 3.33

Answers

(a)

The probability of a typo being caught is

p_{1} + p_{2} - p_{1} p_{2}

. Then,

P (X = k) = (\binom{n}{k}) {(p_{1} + p_{2} - p_{1} p_{2})}^{k} {(1 - (p_{1} + p_{2} - p_{1} p_{2}))}^{n - k}

(b)

When we know the total number of caught typos in advance, the typos caught by the first proofreader are no longer independent. For example, if we know that first proofreader has caught the first

t

typos, and the total number of caught typos is

t

, then the probability of the first proofreader catching subsequent typos is

0

, since the total number of caught typos was

t

Thus, we employ a Hypergeometric distribution. Since $p_{1} = p_{2}$ , all $(\binom{2 n}{t})$ $t$ -tuples of caught typos are equally likely. Hence,

P (X_{1} = k | X_{1} + X_{2} = t) = \frac{(\binom{n}{k}) (\binom{n}{t - k})}{(\binom{2 n}{t})}

fifthist

2021-12-05 00:00

Exercise 3.33

Answers

Comments

Add answer