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1 year ago

You and a friend are playing a game by tossing two coins. If both coins land

Mathematics

2 answers:

Novosadov [1.4K]1 year ago

3 0

c. No. You have a probability of winning, while your friend has a
probability of winning.
Or it’s A

klio [65]1 year ago

3 0

Answer:

Step-by-step explanation:

Let the random variable X be the number of heads that come up when a fair coin is tossed 4 times. Then X∼B(4,12). We want there to be exactly two heads (forcing the other two tosses to be tails), so P(X=2)=(42)(12)2(12)2=38.

You might be interested in

Write a verbal expression for 6m-2

katrin2010 [14]

Answer: The difference of 6 times a number m and 2

Step-by-step explanation:

The way to write this expression is:

The difference of 6 times a number m and 2.

5 0

1 year ago

Choose all sequences of transformations that produce the same image of a given figure.

elena-14-01-66 [18.8K]

Your answer: a reflection across the y-axis followed by a clockwise rotation 90° about the origin. A clockwise rotation 90° about the origin followed by a reflection across the x-axis. A counter-clockwise rotation 90° about the origin followed by a reflection across the y-axis. A reflection across the x-axis followed by a counter-clockwise rotation 90° about the origin.

Hope this helps <3

Stay safe

Stay Warm

-Carrie

Ps. It would mean a lot if you marked this brainliest

5 0

1 year ago

A class with n kids lines up for recess. The order in which the kids line up is random with eachordering being equally likely. T

Elis [28]

Answer:

A) P(Betty is first in line and mary is last) = P(B₁) + P(Mₙ) - (P(B₁) × P(Mₙ/B₁))

B) The method used is Relative frequency approach.

Step-by-step explanation:

From the question, we are told a sample of n kids line up for recess.

Now, the order in which they line up is random with each ordering being equally likely. Thus, this means that the probability of each kid to take a position is n(total of kids/positions).

Since we are being asked about 3 kids from the class, let's assign a letter to each kid:

J: John

B: Betty

M: Mary

A) Now, we want to find the probability that Betty is first in line or Mary is last in line.

In this case, the events are not mutually exclusive, since it's possible that "Betty is first but Mary is not last" or "Mary is last but Betty is not first" or "Betty is the first in line and Mary is last". Thus, there is an intersection between them and the probability is symbolized as;

P(B₁ ∪ Mₙ) = P(B₁) + P (Mₙ) - P(B₁ ∩ Mₙ) = P(B₁) + P(Mₙ) - (P(B₁) × P(Mₙ/B₁))

Where;

The suffix 1 refers to the first position while the suffix n refers to the last position.

Also, P(B₁ ∩ Mₙ) = P(B₁) × P(Mₙ/B₁)

This is because the events "Betty" and "Mary" are not independent since every time a kid takes his place the probability of the next one is affected.

B) The method used is Relative frequency approach.

In this method, the probabilities are usually assigned on the basis of experimentation or historical data.

For example, If A is an event we are considering, and we assume that we have performed the same experiment n times so that n is the number of times A could have occurred.

Also, let n_A be the number of times that A did occur.

Now, the relative frequency would be written as (n_A)/n.

Thus, in this method, we will define P(A) as:

P(A) = lim:n→∞[(n_A)/n]

7 0

2 years ago

The probability of a train arriving on time and leaving on time is 0.8. The probability that the train arrives on time and leave

kkurt [141]

Answer:

0.9524

Step-by-step explanation:

Note enough information is given in this problem. I will do a similar problem like this. The problem is:

The Probability of a train arriving on time and leaving on time is 0.8.The probability of the same train arriving on time is 0.84. The probability of the same train leaving on time is 0.86.Given the train arrived on time, what is the probability it will leave on time?

Solution:

This is conditional probability.

Given:

Probability train arrive on time and leave on time = 0.8
Probability train arrive on time = 0.84
Probability train leave on time = 0.86

Now, according to conditional probability formula, we can write:

$P(Leave \ on \ time | arrive \ on \ time)$ = P(arrive ∩ leave) / P(arrive)

Arrive ∩ leave means probability of arriving AND leaving on time, that is given as "0.8"

and

P(arrive) means probability arriving on time given as 0.84, so:

0.8/0.84 = 0.9524

This is the answer.

5 0

1 year ago

What is the justification for each step in the solution of the equation?

podryga [215]

Here are the following justifications:

Given: -2 (x-1) +12=4x-6

Distribute property = -2x+2+12=4x-6

Combine like terms= -2x+14=4x-6

Transpose = -2x = 4x – 20

Addition or Subtraction Property of Equality = -6x = -20

Division property of equality = x = 20/6

Simplify = x = 10/3

7 0

1 year ago