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3 years ago

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Suppose that a box contains one fair coin (Heads and Tails are equally likely) and one coin with Heads on each side. Suppose fur

ther that someone else selects one of the two coins at random, tosses it three times, and gets the outcome {Heads, Heads, Heads}. In this problem, you will answer the following question in several parts: What is the probability that the chosen coin was the fair coin

1 answer:

stealth61 [152]3 years ago

4 0

Using conditional probability, it is found that there is a 0.1 = 10% probability that the chosen coin was the fair coin.

Conditional Probability

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.
$P(A \cap B)$ is the probability of both A and B happening.
P(A) is the probability of A happening.

In this problem:

Event A: Three heads.
Event B: Fair coin.

The probability associated with 3 heads are:

$0.5^3 = 0.125$ out of 0.5(fair coin).
1 out of 0.5(biased).

Hence:

$P(A) = 0.125 + 0.5 = 0.625$

The probability of 3 heads and the fair coin is:

$P(A \cap B) = 0.5(0.125) = 0.0625$

Then, the conditional probability is:

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.0625}{0.625} = 0.1$

0.1 = 10% probability that the chosen coin was the fair coin.

A similar problem is given at brainly.com/question/14398287

You might be interested in

INT1 4.1.1 Line of Best Fit (classwork)

Lorico [155]

Answer:

a. 2401.06

b. 37.54%

c. 56.3%

Step-by-step explanation:

hopefully this is right

*note: I think you forgot to convert 4-2 back into yards.

a. Robbie's field of view to the North end includes parts not on the football field. so to find the area of the football field he can see, we need to find:

total area of what Robbie sees - area of non football field Robbie sees

what you shaded represents the total of what Robbie sees. it's a triangle. area of a triangle is 1/2(b)(h) where h is distance away and b is width of view.

total area = 1/2(170)(30.92) = 2628.56 yd

area of non football field (pipe to South end)

= 1/2(50)(9.1) = 227.5

so 2628.56 - 227.5 = 2401.06

b. area found in part a / total area of football field

2401.06 / (120*53.3) = .3754

.3754 * 100 = 37.54%

c. the chance of Robbie seeing the touchdown depends on how much of the (North) endzone he can see.

area of north endzone is

10 * 53.3 = 533.

area Robbie sees in endzone is

2628.56 - 2328.48 = 300.08

(found by total area Robbie sees - area of non endzone Robbie sees)

300.08 / 533 = 0.563

= 56.3%

7 0

3 years ago

How do you use a model to show 3.1 = 3.10

n200080 [17]

Because every number with the decimal point have a zero infinit so it's not matter they're are same

5 0

3 years ago

The pressure and volume of an expanding gas are related by the formula PV^b=C, where b and C are constants (this holds in adiaba

bazaltina [42]

The dP/dt of the adiabatic expansion is -42/11 kPa/min

<h3>How to calculate dP/dt in an adiabatic expansion?</h3>

An adiabatic process is a process in which there is no exchange of heat from the system to its surrounding neither during expansion nor during compression

Given b=1.5, P=7 kPa, V=110 cm³, and dV/dt=40 cm³/min

PVᵇ = C

Taking logs of both sides gives:

ln P + b ln V = ln C

Taking partial derivatives gives:

$\frac{1}{P}\frac{∂P}{∂t} + \frac{b}{V}\frac{∂V}{∂t} = 0$

Substitutituting the values b, P, V and dV/dt into the derivative above:

1/7 x dP/dt + 1.5/110 x 40 = 0

1/7 x dP/dt + 6/11 = 0

1/7 x dP/dt = - 6/11

dP/dt = - 6/11 x 7

dP/dt = -42/11 kPa/min

Therefore, the value of dP/dt is -42/11 kPa/min

Learn more about adiabatic expansion on:

brainly.com/question/6966596

#SPJ1

4 0

1 year ago

Round 8.876 to the nearest whole number

Dmitriy789 [7]

8.876 to the nearest whole number
8.876 can be rounded to 9 because if we have a number greater than 5 we round to next number i.e is 9 but if we have a number smaller than 5 then we round to the previous number that is 8.

So your answer is 9

6 0

3 years ago

Read 2 more answers

Write the equation of the line that passes through (−3,1) and (2,−1) in slope-intercept form

Alex787 [66]

Answer:

$y=-\frac{2}{5}x-\frac{1}{5}$

Step-by-step explanation:

The equation of a line is y = mx + b

Where:

m is the slope

b is the y-intercept

First, let's find what m is, the slope of the line.

Let's call the first point you gave, (-3,1), point #1, so the x and y numbers given will be called x1 and y1.

Also, let's call the second point you gave, (2,-1), point #2, so the x and y numbers here will be called x2 and y2.

Now, just plug the numbers into the formula for m above, like this:

$m = -\frac{2}{5}$

So, we have the first piece to finding the equation of this line, and we can fill it into y=mx+b like this:

$y=-\frac{2}{5}x + b$

Now, what about b, the y-intercept?

To find b, think about what your (x,y) points mean:

(-3,1). When x of the line is -3, y of the line must be 1.
(2,-1). When x of the line is 2, y of the line must be -1.

Now, look at our line's equation so far: $y=-\frac{2}{5}x + b$ . $b$ is what we want, the - $-\frac{2}{5}$ is already set and x and y are just two 'free variables' sitting there. We can plug anything we want in for x and y here, but we want the equation for the line that specfically passes through the two points (-3,1) and (2,-1).

So, why not plug in for x and y from one of our (x,y) points that we know the line passes through? This will allow us to solve for b for the particular line that passes through the two points you gave!

You can use either (x,y) point you want. The answer will be the same:

(-3,1). y = mx + b or $1=-\frac{2}{5} * -3 + b$ , or solving for b: $b = 1-(-\frac{2}{5})(-3)$ . $b = -\frac{1}{5}$ .
(2,-1). y = mx + b or $-1=-\frac{2}{5} * 2 + b$ , or solving for b: $b = 1-(-\frac{2}{5})(2)$ . $b = -\frac{1}{5}$ .

See! In both cases, we got the same value for b. And this completes our problem.

The equation of the line that passes through the points (-3,1) and (2,-1) is $y=-\frac{2}{5}x-\frac{1}{5}$

8 0

3 years ago