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Lesechka [4]
3 years ago
10

Four cards are shuffled and placed face down in front of you. Their faces (hidden) display the four elements: water, earth, wind

, and fire. You are to turn the cards over one card at a time until y ou either win or lose. You win if both the water and earth cards are turned over. You lose if the fire card is turn ed over. What is the probability of winning?
Mathematics
1 answer:
Licemer1 [7]3 years ago
5 0

Answer:

probability of winning is\frac{2}{3}

Step-by-step explanation:

As it is given , 4 cards are available,  there are 4!  mean total possible way of ordering is 24  

following are the cases of loosing

1) first card is fire.

Thus there are 3! = mean 6 possible ways for particular ordering  to occur.

2) If wind is first card and Fire is second. Then, There are 2! = 2 ways to occur.

Hence, the probabiIity of losing is given as  = \frac{6+2}{24}=\frac{1}{3}

therefore probality of winning is 1 -\frac{1}{3}

probality of winning is\frac{2}{3}

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Imagine that you would like to purchase a $275,000 home. Using 20% as
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Answer:

The mortgage chosen is option A;

15-year mortgage term with a 3% interest rate because it has the lowest total amount paid over the loan term of $270,470

Step-by-step explanation:

The details of the home purchase are;

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The mode of purchase of the home = Mortgage

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The amount used as down payment for the loan = $55,000

The principal of the mortgage borrowed, P = The price of the house - The down payment

∴ P = $275,000 - 20/100 × $275,000 = $275,000 - $55,000 = $220,000

The principal of the mortgage, P = $220,000

The formula for the total amount paid which is the cost of the loan is given as follows;

Outstanding \ Loan \ Balance = \dfrac{P \cdot \left[\left(1+\dfrac{r}{12} \right)^n -  \left(1+\dfrac{r}{12} \right)^m \right] }{1 - \left(1+\dfrac{r}{12} \right)^n }

The formula for monthly payment on a mortgage, 'M', is given as follows;

M = \dfrac{P \cdot \left(\dfrac{r}{12} \right) \cdot \left(1+\dfrac{r}{12} \right)^n }{\left(1+\dfrac{r}{12} \right)^n - 1}

A. When the mortgage term, t = 15-years,

The interest rate, r = 3%

The number of months over which the loan is payed, n = 12·t

∴ n = 12 months/year × 15 years = 180 months

n = 180 months

The monthly payment, 'M', is given as follows;

M =

The total amount paid over the loan term = Cost of the mortgage

Therefore, we have;

220,000*0.05/12*((1 + 0.05/12)^360/( (1 + 0.05/12)^(360) - 1)

M = \dfrac{220,000 \cdot \left(\dfrac{0.03}{12} \right) \cdot \left(1+\dfrac{0.03}{12} \right)^{180} }{\left(1+\dfrac{0.03}{12} \right)^{180} - 1}  \approx 1,519.28

The minimum monthly payment for the loan, M ≈ $1,519.28

The total amount paid over loan term, A = n × M

∴ A ≈ 180 × $1,519.28 = $273,470

The total amount paid over loan term, A ≈ $270,470

B. When t = 20 year and r = 6%, we have;

n = 12 × 20 = 240

\therefore M = \dfrac{220,000 \cdot \left(\dfrac{0.06}{12} \right) \cdot \left(1+\dfrac{0.06}{12} \right)^{240} }{\left(1+\dfrac{0.06}{12} \right)^{240} - 1}  \approx 1,576.15

The total amount paid over loan term, A = 240 × $1,576.15 ≈ $378.276

The monthly payment, M = $1,576.15

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n = 12 × 30 = 360

\therefore M = \dfrac{220,000 \cdot \left(\dfrac{0.05}{12} \right) \cdot \left(1+\dfrac{0.05}{12} \right)^{360} }{\left(1+\dfrac{0.05}{12} \right)^{360} - 1}  \approx 1,181.01

The total amount paid over loan term, A = 360 × $1,181.01 ≈ $425,163

The monthly payment, M ≈ $1,181.01

The mortgage to be chosen is the mortgage with the least total amount paid over the loan term so as to reduce the liability

Therefore;

The mortgage chosen is option A which is a 15-year mortgage term with a 3% interest rate;

The total amount paid over the loan term = $270,470

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