4.3 Problems on independence of events (Page 5/5)

Applied probability Page 5 / 5

Ten numbers are produced by a random number generator. What is the probability four or more are less than or equal to 0.63?

$P 1 = c b i n o m (10, 0.63, 4) = 0.9644$

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A player rolls a pair of dice five times. She scores a “hit” on any throw if she gets a 6 or 7. She wins iff she scores an odd number of hits in the five throws. What is the probability a player wins on any sequence of five throws?Suppose she plays the game 20 successive times. What is the probability she wins at least 10 times? What is the probability she wins more than half the time?

Each roll yields a hit with probability $p = \frac{6}{36} + \frac{5}{36} = \frac{11}{36}$ .

PW = sum(ibinom(5,11/36,[1 3 5]))PW =  0.4956
P2 = cbinom(20,PW,10)P2 =  0.5724
P3 = cbinom(20,PW,11)P3 =  0.3963

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Erica and John spin a wheel which turns up the integers 0 through 9 with equal probability. Results on various trials are independent. Each spins the wheel10 times. What is the probability Erica turns up a seven more times than does John?

$P = i b i n o m (10, 0.1, 0 : 9) * c b i n o m (10, 0.1, 1 : 10)' = 0.3437$

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Erica and John play a different game with the wheel, above. Erica scores a point each time she gets an integer 0, 2, 4, 6, or 8. John scores a pointeach time he turns up a 1, 2, 5, or 7. If Erica spins eight times; John spins 10 times. What is the probability John makes more points than Erica?

$P = i b i n o m (8, 0.5, 0 : 8) * c b i n o m (10, 0.4, 1 : 9)' = 0.4030$

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A box contains 100 balls; 30 are red, 40 are blue, and 30 are green. Martha and Alex select at random, with replacement and mixing after each selection. Alex has asuccess if he selects a red ball; Martha has a success if she selects a blue ball. Alex selects seven times and Martha selects five times. What is the probabilityMartha has more successes than Alex?

$P = i b i n o m (7, 0.3, 0 : 4) * c b i n o m (5, 0.4, 1 : 5)' = 0.3613$

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Two players roll a fair die 30 times each. What is the probability that each rolls the same number of sixes?

$P = s u m (i b i n o m (30, 1 / 6, 0 : 30) .^2) = 0.1386$

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A warehouse has a stock of n items of a certain kind, r of which are defective. Two of the items are chosen at random, without replacement. What is theprobability that at least one is defective? Show that for large n the number is very close to that for selection with replacement, which corresponds to twoBernoulli trials with pobability $p = r / n$ of success on any trial.

P 1 = \frac{r}{n} • \frac{r - 1}{n - 1} + \frac{r}{n} • \frac{n - r}{n - 1} + \frac{n - r}{n} • \frac{r}{n - 1} = \frac{(2 n - 1) r - r^{2}}{n (n - 1)}

P 2 = 1 - {(\frac{r}{n})}^{2} = \frac{2 n r - r^{2}}{n^{2}}

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A coin is flipped repeatedly, until a head appears. Show that with probability one the game will terminate.

The probability of not terminating in n trials is q ⁿ .

Let $N =$ event never terminates and $N_{k} =$ event does not terminate in k plays. Then $N \subset N_{k}$ for all k implies $0 \leq P (N) \leq P (N_{k}) = 1 / 2^{k}$ for all k . We conclude $P (N) = 0$ .

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Two persons play a game consecutively until one of them is successful or there are ten unsuccesful plays. Let E _i be the event of a success on the i th play of the game. Suppose ${E_{i} : 1 \leq i}$ is an independent class with $P (E_{i}) = p_{1}$ for i odd and $P (E_{i}) = p_{2}$ for i even. Let A be the event the first player wins, B be the event the second player wins, and C be the event that neither wins.

Express $A, B$ , and C in terms of the E _i .
Determine $P (A)$ , $P (B)$ , and $P (C)$ in terms of p ₁ , p ₂ , $q_{1} = 1 - p_{1}$ , and $q_{2} = 1 - p_{2}$ . Obtain numerical values for the case $p_{1} = 1 / 4$ and $p_{2} = 1 / 3$ .
Use appropriate facts about the geometric series to show that $P (A) = P (B)$ iff $p_{1} = p_{2} / (1 + p_{2})$ .
Suppose $p_{2} = 0.5$ . Use the result of part (c) to find the value of p ₁ to make $P (A) = P (B)$ and then determine $P (A)$ , $P (B)$ , and $P (C)$ .

$C = ⋂_{i = 1}^{10} E_{i}^{c}$ .
$A = E_{1} ⋁ E_{1}^{c} E_{2}^{c} E_{3} ⋁ E_{1}^{c} E_{2}^{c} E_{3}^{c} E_{4}^{c} E_{5} ⋁ E_{1}^{c} E_{2}^{c} E_{3}^{c} E_{4}^{c} E_{5}^{c} E_{6}^{c} E_{7} ⋁ E_{1}^{c} E_{2}^{c} E_{3}^{c} E_{4}^{c} E_{5}^{c} E_{6}^{c} E_{7}^{c} E_{8}^{c} E_{9}$

$B = E_{1}^{c} E_{2} ⋁ E_{1}^{c} E_{2}^{c} E_{3}^{c} E_{4} ⋁ E_{1}^{c} E_{2}^{c} E_{3}^{c} E_{4}^{c} E_{5}^{c} E_{6} ⋁ E_{1}^{c} E_{2}^{c} E_{3}^{c} E_{4}^{c} E_{5}^{c} E_{6}^{c} E_{7}^{c} E_{8} ⋁ E_{1}^{c} E_{2}^{c} E_{3}^{c} E_{4}^{c} E_{5}^{c} E_{6}^{c} E_{7}^{c} E_{8}^{c} E_{9}^{c} E_{10}$
$P (A) = p_{1} [1 + q_{1} q_{2} + {(q_{1} q_{2})}^{2} + {(q_{1} q_{2})}^{3} + {(q_{1} q_{2})}^{4}] = p_{1} \frac{1 - {(q_{1} q_{2})}^{5}}{1 - q_{1} q_{2}}$

$P (B) = q_{1} p_{2} \frac{1 - {(q_{1} q_{2})}^{5}}{1 - q_{1} q_{2}} P (C) = {(q_{1} q_{2})}^{5}$
For $p_{1} = 1 / 4, p_{2} = 1 / 3$ , we have $q_{1} q_{2} = 1 / 2$ and $q_{1} p_{2} = 1 / 4$ . In this case
$P (A) = \frac{1}{4} • \frac{31}{16} = 31 / 64 = 0.4844 = P (B), P (C) = 1 / 32$
Note that $P (A) + P (B) + P (C) = 1$ .
$P (A) = P (B)$ iff $p_{1} = q_{1} p_{2} = (1 - p_{1}) p_{2}$ iff $p_{1} = p_{2} / (1 + p_{2})$ .
$p_{1} = 0.5 / 1.5 = 1 / 3$

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Three persons play a game consecutively until one achieves his objective. Let E _i be the event of a success on the i th trial, and suppose ${E_{i} : 1 \leq i}$ is an independent class, with $P (E_{i}) = p_{1}$ for $i = 1, 4, 7, \dots$ , $P (E_{i}) = p_{2}$ for $i = 2, 5, 8, \dots$ , and $P (E_{i}) = p_{3}$ for $i = 3, 6, 9, \dots$ . Let $A, B, C$ be the respective events the first, second, and third player wins.

Express $A, B$ , and C in terms of the E _i .
Determine the probabilities in terms of $p_{1}, p_{2}, p_{3}$ , then obtain numerical values in the case $p_{1} = 1 / 4$ , $p_{2} = 1 / 3$ , and $p_{3} = 1 / 2$ .

- $A = E_{1} ⋁ ⋁_{k = 1}^{\infty} ⋂_{i = 1}^{3 k} E_{i}^{c} E_{3 k + 1}$
- $B = E_{1}^{c} E_{2} ⋁ ⋁_{k = 1}^{\infty} ⋂_{i = 1}^{3 k + 1} E_{i}^{c} E_{3 k + 2}$
- $C = E_{1}^{c} E_{2}^{c} E_{3} ⋁ ⋁_{k = 1}^{\infty} ⋂_{i = 1}^{3 k + 2} E_{i}^{c} E_{3 k + 3}$
- $P (A) = p_{1} \sum_{k = 0}^{\infty} {(q_{1} q_{2} q_{3})}^{k} = \frac{p_{1}}{1 - q_{1} q_{2} q_{3}}$
- $P (B) = \frac{q_{1} p_{2}}{1 - q_{1} q_{2} q_{3}}$
- $P (C) = \frac{q_{1} q_{2} p_{3}}{1 - q_{1} q_{2} q_{3}}$
- For $p_{1} = 1 / 4$ , $p_{2} = 1 / 3$ , $p_{3} = 1 / 2$ , $P (A) = P (B) = P (C) = 1 / 3$ .

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What is the probability of a success on the i th trial in a Bernoulli sequence of n component trials, given there are r successes?

$P (A_{r n} E_{i}) = p C (n - 1, r - 1) p^{r - 1} q^{n - r}$ and $P (A_{r n}) = C (n, r) p^{r} q^{n - r}$ .

Hence $P (E_{i} | A_{A} r n) = C (n - 1, r - 1) / C (n, r) = r / n$ .

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A device has N similar components which may fail independently, with probability p of failure of any component. The device fails if one or more of the components fails. In the event of failure of the device, the components are testedsequentially.

What is the probability the first defective unit tested is the n th, given one or more components have failed?
What is the probability the defective unit is the n th, given that exactly one has failed?
What is the probability that more than one unit has failed, given that the first defective unit is the n th?

Let $A_{1} =$ event one failure, $B_{1} =$ event of one or more failures, $B_{2} =$ event of two or more failures, and $F_{n} =$ the event the first defective unit found is the n th.

$F_{n} \subset B_{1}$ implies $P (F_{n} | B_{1}) = P (F_{n}) / P (B_{1}) = \frac{q^{n - 1} p}{1 - q^{N}}$
$P (F_{n} | A_{1}) = \frac{P (F_{n} A_{1})}{P (A_{1})} = \frac{q^{n - 1} p q^{N - n}}{N p q^{N - 1}} = \frac{1}{N}$
(see [link] )
Since probability not all from n th are good is $1 - q^{N - n}$ ,
$P (B_{2} | F_{n}) = \frac{P (B_{2} F_{n})}{P (F_{n})} = \frac{q^{n - 1} p (1 - Q^{N - 1})}{q^{n - 1} p} = 1 - q^{N - n}$

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Source: OpenStax, Applied probability. OpenStax CNX. Aug 31, 2009 Download for free at http://cnx.org/content/col10708/1.6

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