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mote1985 [20]
1 year ago
14

A survey found that 73% of adults have a landline at their residence (event A); 83% have a cell phone (event B). It is known tha

t 2% of adults have neither a cell phone nor a landline. 3. What is the probability that an adult selected at random has both a landline and a cell phone? A. 0.58 B. 0.98 C. 0.6059 D. None of these Work: 4. Given an adult has a cell phone, what is the probability he does not have a landline?
A. 0.27
B. 0.25
C. 0.3012
D. None of these
Mathematics
1 answer:
4vir4ik [10]1 year ago
4 0

Answer:

3. What is the probability that an adult selected at random has both a landline and a cell phone?

A. 0.58

4. Given an adult has a cell phone, what is the probability he does not have a landline?

C. 0.3012

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that an adult has a landline at his residence.

B is the probability that an adult has a cell phone.

C is the probability that a mean is neither of those.

We have that:

A = a + (A \cap B)

In which a is the probability that an adult has a landline but not a cell phone and A \cap B is the probability that an adult has both of these things.

By the same logic, we have that:

B = b + (A \cap B)

The sum of all the subsets is 1:

a + b + (A \cap B) + C = 1

2% of adults have neither a cell phone nor a landline.

This means that C = 0.02.

73% of adults have a landline at their residence (event A); 83% have a cell phone (event B)

So A = 0.73, B = 0.83.

What is the probability that an adult selected at random has both a landline and a cell phone?

This is A \cap B.

We have that A = 0.73. So

A = a + (A \cap B)

a = 0.73 - (A \cap B)

By the same logic, we have that:

b = 0.83 - (A \cap B).

So

a + b + (A \cap B) + C = 1

0.73 - (A \cap B) + 0.83 - (A \cap B) + (A \cap B) + 0.02 = 1

(A \cap B) = 0.75 + 0.83 - 1 = 0.58

So the answer for question 3 is A.

4. Given an adult has a cell phone, what is the probability he does not have a landline?

83% of the adults have a cellphone.

We have that

b = B - (A \cap B) = 0.83 - 0.58 = 0.25

25% of those do not have a landline.

So P = \frac{0.25}{0.83} = 0.3012

The answer for question 4 is C.

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Hello!

To find how much money total will be in the account after four years, you need to use the function, y = a(1 + r)^{t}, as seen above. In this function, a is the starting value, r is the interest rate, and t is the amount of time (in years).

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1 year ago
Two wires with lengths of 448 cm and 616 cm are to be cut into pieces of all the same length without a remainder. Find the great
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Answer:

56 cm

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We need ro find the GCD of these numbers. In finding the GCD, we list the multiples of the number, beginning with the smallest number. Here, the

The factors of 616=2*2*2*7*11

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Jordan wants to play a basketball game at a carnival. The game costs the player $5 dollar sign, 5 to play, and the player gets t
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Answer:

The expected value of Jordan gains is -1 dollar.

Step-by-step explanation:

Consider the following random variables. X := #of shots that Jordan makes. Then, we can can define the random variable Y of the earnings of Jordan in a game as follows

Y = 5 if X=2 (since he gets 10, but invested 5),  Y=0 if X=1(since he gets the 5 back) and Y=-5 if X=0(since he doesn't get the money back). Then, in this case, we can define the probability as follows.

P(Y=5) = P(X=2), P(Y=0) = P(X=1), P(Y=-5)= P(X=0).

By definition, the expected value of Y is given by

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We only need to calculate the probabilities for X. In this case, we can consider each shot independt from each other. Then, we can consider X to be distributed as a binomial random variable with n=2 trials and p=0.4 of success (since he has a 40% chance of winning).

Then, by definition

P(X=k) = \binom{n}{k}p^k(1-p)^{n-k} = \binom{2}{k}0.4^{k}0.6^{2-k}

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