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Yuliya22 [10]
3 years ago
15

A mutual fund company offers its customers a variety of funds: a money-market fund, three different bond funds (short, intermedi

ate, and long-term), two stock funds (moderate and high-risk), and a balanced fund. Among customers who own shares in just one fund, the percentages of customers in the different funds are as follows:
Money market 20%
Short-term bond 15%
Intermediate-term bond 11%
Long-term bond 5%
High-risk stock 18%
Moderate-risk stock 24%
Balanced fund 7%

A customer who owns shares in just one fund is to be selected at random.

Required:
a. What is the probability that the selected individual owns shares in the balanced fund?
b. What is the probability that the individual owns shares in a bond fund?
c. What is the probability that the selected individual does not own shares in a stock fund?
Mathematics
1 answer:
bagirrra123 [75]3 years ago
3 0

Answer:

a. 7% probability that the selected individual owns shares in the balanced fund.

b. 31% probability that the individual owns shares in a bond fund.

c. 58% probability that the selected individual does not own shares in a stock fund

Step-by-step explanation:

a. What is the probability that the selected individual owns shares in the balanced fund?

7% of the individuals owns shares in the balanced fund, so 7% probability that the selected individual owns shares in the balanced fund.

b. What is the probability that the individual owns shares in a bond fund?

Bond funds (short, intermediate, and long-term):

15%(short-term) + 11%(intermediate-term) + 5%(long-term)

15%+11%+5% = 31%

31% probability that the individual owns shares in a bond fund.

c. What is the probability that the selected individual does not own shares in a stock fund?

Anything but moderate and high risk stock%.

The sum of all probabilities is 100%.

18% + 24% = 42% own stock fund.

100 - 42 = 58%

58% probability that the selected individual does not own shares in a stock fund

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Here, the left limit is 2 and the right limit is 1/2. The limits are different, which is why the function is discontinuous at x=-1.

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Use the change of base formula to evaluate log5 44. Then convert log5 44 to a logarithm in base 2. Round to the nearest thousand
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A bag contains two six-sided dice: one red, one green. The red die has faces numbered 1, 2, 3, 4, 5, and 6. The green die has fa
gayaneshka [121]

Answer:

the probability the die chosen was green is 0.9

Step-by-step explanation:

Given that:

A bag contains two six-sided dice: one red, one green.

The red die has faces numbered 1, 2, 3, 4, 5, and 6.

The green die has faces numbered 1, 2, 3, 4, 4, and 4.

From above, the probability of obtaining 4 in a single throw of a fair die is:

P (4  | red dice) = \dfrac{1}{6}

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As the die is selected randomly; the probability of the first die must be equal to the probability of the second die = \dfrac{1}{2}

The probability of two 1's and two 4's in the first dice can be calculated as:

= \begin {pmatrix}  \left \begin{array}{c}4\\2\\ \end{array} \right  \end {pmatrix} \times  \begin {pmatrix} \dfrac{1}{6}  \end {pmatrix}  ^4

= \dfrac{4!}{2!(4-2)!} ( \dfrac{1}{6})^4

= \dfrac{4!}{2!(2)!} \times ( \dfrac{1}{6})^4

= 6 \times ( \dfrac{1}{6})^4

= (\dfrac{1}{6})^3

= \dfrac{1}{216}

The probability of two 1's and two 4's in the second  dice can be calculated as:

= \begin {pmatrix}  \left \begin{array}{c}4\\2\\ \end{array} \right  \end {pmatrix} \times  \begin {pmatrix} \dfrac{1}{6}  \end {pmatrix}  ^2  \times  \begin {pmatrix} \dfrac{3}{6}  \end {pmatrix}  ^2

= \dfrac{4!}{2!(2)!} \times ( \dfrac{1}{6})^2 \times  ( \dfrac{3}{6})^2

= 6 \times ( \dfrac{1}{6})^2 \times  ( \dfrac{3}{6})^2

= ( \dfrac{1}{6}) \times  ( \dfrac{3}{6})^2

= \dfrac{9}{216}

∴

The probability of two 1's and two 4's in both dies = P( two 1s and two 4s | first dice ) P( first dice ) + P( two 1s and two 4s | second dice ) P( second dice )

The probability of two 1's and two 4's in both die = \dfrac{1}{216} \times \dfrac{1}{2} + \dfrac{9}{216} \times \dfrac{1}{2}

The probability of two 1's and two 4's in both die = \dfrac{1}{432}  + \dfrac{1}{48}

The probability of two 1's and two 4's in both die = \dfrac{5}{216}

By applying  Bayes Theorem; the probability that the die was green can be calculated as:

P(second die (green) | two 1's and two 4's )  = The probability of two 1's and two 4's | second dice)P (second die) ÷ P(two 1's and two 4's in both die)

P(second die (green) | two 1's and two 4's )  = \dfrac{\dfrac{1}{2} \times \dfrac{9}{216}}{\dfrac{5}{216}}

P(second die (green) | two 1's and two 4's )  = \dfrac{0.5 \times 0.04166666667}{0.02314814815}

P(second die (green) | two 1's and two 4's )  = 0.9

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kolezko [41]
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