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Gnom [1K]
3 years ago
13

In a large school, it was found that 71% of students are taking a math class, 77% of student are taking an English class, and 58

% of students are taking both. Find the probability that a randomly selected student is taking a math class or an English class. Write your answer as a decimal, and round to 2 decimal places if necessary.
Mathematics
1 answer:
Anni [7]3 years ago
8 0

Answer:

P (Math or English) = 0.90

Step-by-step explanation:

* Lets study the meaning of or , and on probability

- The use of the word or means that you are calculating the probability

  that either event A or event B happened

- Both events do not have to happen

- The use of the word and, means that both event A and B have to

  happened

* The addition rules are:

# P(A or B) = P(A) + P(B) ⇒ mutually exclusive (events cannot happen

  at the same time)

# P(A or B) = P(A) + P(B) - P(A and B) ⇒ non-mutually exclusive (if they  

   have at least one outcome in common)

- The union is written as A ∪ B or “A or B”.  

- The Both is written as A ∩ B or “A and B”

* Lets solve the question

- The probability of taking Math class 71%

- The probability of taking English class 77%

- The probability of taking both classes is 58%

∵ P(Math) = 71% = 0.71

∵ P(English) = 77% = 0.77

∵ P(Math and English) = 58% = 0.58

- To find P(Math or English) use the rule of non-mutually exclusive

∵ P(A or B) = P(A) + P(B) - P(A and B)

∴ P(Math or English) = P(Math) + P(English) - P(Math and English)

- Lets substitute the values of P(Math) , P(English) , P(Math and English)

 in the rule

∵ P(Math or English) = 0.71 + 0.77 - 0.58 ⇒ simplify

∴ P(Math or English) =  0.90

* P(Math or English) = 0.90

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Mamont248 [21]

Answer:

The probability that these four will contain at least two more high-risk drivers than low-risk drivers is 0.0488.

Step-by-step explanation:

Denote the different kinds of drivers as follows:

L = low-risk drivers

M = moderate-risk drivers

H = high-risk drivers

The information provided is:

P (L) = 0.50

P (M) = 0.30

P (H) = 0.20

Now, it given that the insurance company writes four new policies for adults earning their first driver’s license.

The combination of 4 new drivers that satisfy the condition that there are at least two more high-risk drivers than low-risk drivers is:

S = {HHHH, HHHL, HHHM, HHMM}

Compute the probability of the combination {HHHH} as follows:

P (HHHH) = [P (H)]⁴

                = [0.20]⁴

                = 0.0016

Compute the probability of the combination {HHHL} as follows:

P (HHHL) = {4\choose 1} × [P (H)]³ × P (L)

               = 4 × (0.20)³ × 0.50

               = 0.016

Compute the probability of the combination {HHHM} as follows:

P (HHHL) = {4\choose 1} × [P (H)]³ × P (M)

               = 4 × (0.20)³ × 0.30

               = 0.0096

Compute the probability of the combination {HHMM} as follows:

P (HHMM) = {4\choose 2} × [P (H)]² × [P (M)]²

                 = 6 × (0.20)² × (0.30)²

                 = 0.0216

Then the probability that these four will contain at least two more high-risk drivers than low-risk drivers is:

P (at least two more H than L) = P (HHHH) + P (HHHL) + P (HHHM)

                                                            + P (HHMM)

                                                  = 0.0016 + 0.016 + 0.0096 + 0.0216

                                                  = 0.0488

Thus, the probability that these four will contain at least two more high-risk drivers than low-risk drivers is 0.0488.

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Suppose y varies directly as x write a direct variation equation that relates x and y
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Answer:

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Answer:

Step-by-step explanation:

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multiply (ii) by -1. so a will be eliminated

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Plug in the value of d in equation (i),

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