cross-ied

The number of days a particular stock increases in a given five-day period is uniformly distributed between zero and five inclusive. In a given five-day trading week, what is the probability that the stock will increase exactly three days?

A) 0.333.

B) 0.167.

C) 0.600.

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If the possible outcomes are X0,1,2,3,4,5), then the probability of each of the six outcomes is 1 / 6 = 0.167.

cross-ied

The Night Raiders, an expansion team in the National Indoor Football League, is having a challenging first season with a current win loss record of 0 and 4. However, the team recently signed four new defensive players and one of the team sponsors (who also happens to hold a CFA charter) calculates the probability of the team winning a game at 0.40. Assuming that whether the team wins a game is independent of whether it wins any other game, the probability that the team will win 6 out of the next 10 games is closest to:

A) 0.417.

B) 0.112.

C) 0.350.

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Use the formula for a binomial random variable to calculate the answer to this question. We will define "success" as the team winning a game. The formula is:

p(x) = P(X = x) = [number of ways to choose x from n] × px × (1 - p)n-x,

where [number of ways to choose x from n] = n! / [(n - x)! × x!].
Here, p(x) = P(X = 6) = [10! / (10 − 6)! × 6!] × 0.406 × (1 − 0.40)10-6
= 210.0 × 0.00410 × 0.12960 = 0.11159, or approximately 0.112.
To calculate factorial using your financial calculator: On the TI, factorial is [2nd] ¡→ [x!]. On the HP, factorial is [g] → [n!]. To compute 10! on the TI, enter [10] → [2nd] → [x!] = 3,628,800. On the HP, use [10] → [ENTER] → [g] → [n!].

JustasS

Assume 30% of the CFA candidates have a degree in economics. A random sample of three CFA candidates is selected. What is the probability that none of them has a degree in economics?

A) 0.343.

B) 0.027.

C) 0.900.

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3! / (0!3!) (0.3)0 (0.7)3 = 0.343

JustasS

A casual laborer has a 70% chance of finding work on each day that she reports to the day labor marketplace. What is the probability that she will work three days out of five?

A) 0.6045.

B) 0.3192.

C) 0.3087.

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P(3) = 5! / [(5 – 3)! × 3!] × (0.73) × (0.32) = 0.3087 = 5 →2nd→ nCr → 3 ×  0.343  × 0.09

JustasS

Which of the following could be the set of all possible outcomes for a random variable that follows a binomial distribution?

A) (-1, 0, 1).

B) (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11).

C) (1, 2).

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This reflects a basic property of binomial outcomes. They take on whole number values that must start at zero up to the upper limit n. The upper limit in this case is 11.

JustasS

Which of the following is NOT an assumption of the binomial distribution?

A) The trials are independent.

B) The expected value is a whole number.

C) Random variable X is discrete.

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The expected value is n × p. A simple example shows us that the expected value does not have to be a whole number: n = 5, p = 0.5, n × p = 2.5. The other conditions are necessary for the binomial distribution.

JustasS

For a certain class of junk bonds, the probability of default in a given year is 0.2. Whether one bond defaults is independent of whether another bond defaults. For a portfolio of five of these junk bonds, what is the probability that zero or one bond of the five defaults in the year ahead?

A) 0.0819.

B) 0.4096.

C) 0.7373.

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The outcome follows a binomial distribution where n = 5 and p = 0.2. In this case p(0) = 0.85 = 0.3277 and p(1) = 5 × 0.84 × 0.2 = 0.4096, so P(X=0 or X=1) = 0.3277 + 0.4096.

JustasS

A stock priced at $10 has a 60% probability of moving up and a 40% probability of moving down. If it moves up, it increases by a factor of 1.06. If it moves down, it decreases by a factor of 1/1.06. What is the expected stock price after two successive periods?

A) $10.27.

B) $10.03.

C) $11.24.

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If the stock moves up twice, it will be worth $10 × 1.06 × 1.06 = $11.24. The probability of this occurring is 0.60 × 0.60 = 0.36. If the stock moves down twice, it will be worth $10 × (1/1.06) × (1/1.06) = $8.90. The probability of this occurring is 0.40 × 0.40 = 0.16. If the stock moves up once and down once, it will be worth $10 × 1.06 × (1/1.06) = $10.00. This can occur if either the stock goes up then down or down then up. The probability of this occurring is 0.60 × 0.40 + 0.40 × 0.60 = 0.48. Multiplying the potential stock prices by the probability of them occurring provides the expected stock price: ($11.24 × 0.36) + ($8.90 × 0.16) + ($10.00 × 0.48) = $10.27.

JustasS

A stock priced at $20 has an 80% probability of moving up and a 20% probability of moving down. If it moves up, it increases by a factor of 1.05. If it moves down, it decreases by a factor of 1/1.05. What is the expected stock price after two successive periods?

A) $21.24.

B) $20.05.

C) $22.05.

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If the stock moves up twice, it will be worth $20 × 1.05 × 1.05 = $22.05. The probability of this occurring is 0.80 × 0.80 = 0.64. If the stock moves down twice, it will be worth $20 × (1/1.05) × (1/1.05) = $18.14. The probability of this occurring is 0.20 × 0.20 = 0.04. If the stock moves up once and down once, it will be worth $20 × 1.05 × (1/1.05) = $20.00. This can occur if either the stock goes up then down or down then up. The probability of this occurring is 0.80 × 0.20 + 0.20 × 0.80 = 0.32. Multiplying the potential stock prices by the probability of them occurring provides the expected stock price: ($22.05 × 0.64) + ($18.14 × 0.04) + ($20.00 × 0.32) = $21.24.

JustasS

A stock priced at $100 has a 70% probability of moving up and a 30% probability of moving down. If it moves up, it increases by a factor of 1.02. If it moves down, it decreases by a factor of 1/1.02. What is the probability that the stock will be $100 after two successive periods?

A) 21%.

B) 9%.

C) 42%.

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For the stock to be $100 after two periods, it must move up once and move down once: $100 × 1.02 × (1/1.02) = $100. This can happen in one of two ways: 1) the stock moves up during period one and down during period two; or 2) the stock moves down during period one and up during period two. The probability of either event is 0.70 × 0.30 = 0.21. The combined probability of either event is 2(0.21) = 0.42 or 42%.