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scoundrel [369]
3 years ago
13

A company is bidding on two projects, A and B. The probability that the company wins project A is 0.40 and the probability that

the company wins project B is 0.25. Winning project A and winning project B are independent events. What is the probability that the company wins project A or project B
Mathematics
1 answer:
kondaur [170]3 years ago
3 0

Answer:

Probability that the company wins project A or project B is 0.50.

Step-by-step explanation:

We are given that a company is bidding on two projects, A and B. The probability that the company wins project A is 0.40 and the probability that the company wins project B is 0.25.

Also, Winning project A and winning project B are independent events.

Let the Probability of winning project A = P(A) = 0.40

            Probability of winning project B = P(B) = 0.25

<u>Now, as we know that ;</u>

Probability that the company wins project A or project B = P(A \bigcup B)

              P(A \bigcup B) = P(A) + P(B) -  P(A \bigcap B)

So, we have to find the value of Probability of winning project A and B, i.e;

P(A \bigcap B)

<em>Since, we are given that Winning project A and winning project B are independent events which means when this condition is given then;</em>

                    P(A \bigcap B) = P(A) \times P(B)

                                    = 0.40 \times 0.25 = 0.10

Now, Probability that the company wins project A or project B is given by;

                 P(A \bigcup B) = P(A) + P(B) -  P(A \bigcap B)  

                                  = 0.40 + 0.25 - 0.10

                                  = 0.65 - 0.10 = 0.55

Hence, probability that the company wins project A or project B is 0.50.

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Douglas invests money in two simple interest accounts. He invests three times as much in an account paying 14% as he does in an
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Douglas invested $ 1300 altogether

<em><u>Solution:</u></em>

Let x represent the amount invested in the account paying 14% interest.

Let y represent the amount invested in the account paying 5% interest

He invests three times as much in an account paying 14% as he does in an account paying 5%

Which means,

x = 3y

<em><u>The simple interest is given by formula:</u></em>

S.I = \frac{p \times n \times r}{100}

Where,

"p" is the principal

"n" is the number of years

"r" is the rate of interest

<em><u>He earns $152.75 in interest in one year from both accounts combined</u></em>

Therefore,

Combined S.I = 152.75

n = 1 year

<em><u>Considering the account earning 14% interest:</u></em>

S.I = \frac{3y \times 14 \times 1}{100}\\\\S.I = 0.42y

<em><u>Considering the account earning 5% interest:</u></em>

S.I = \frac{y \times 5 \times 1}{100}\\\\S.I = 0.05y

<em><u>Since, Combined S.I = 152.75</u></em>

Therefore,

0.42y + 0.05y = 152.75

0.47y = 152.75

Divide both sides by 0.47

y = 325

Therefore,

x = 3y

x = 3(325)

x = 975

<em><u>how much did he invest altogether?</u></em>

Amount invested together = x + y = 975 + 325 = 1300

Thus he invested $ 1300 altogether

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