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velikii [3]
3 years ago
8

A litter of puppies consists of black puppies and white puppies. A

Mathematics
1 answer:
BartSMP [9]3 years ago
7 0

We want to see if the first selection is dependent on the second selection, we will see that no, the first selection can not depend on a selection that <u>did not happen yet.</u>

So the situation is:

We have a group of puppies, some are black, some are white.

Let's say that there are N puppies in total, n₁ are black, n₂ are white, such that:

n₁ + n₂ = N

Now we have that the first selection, event A, is a black puppy. The probability is just the quotient between the number of black puppies and the total number of puppies:

q = n₁/N

Then now we have:

  • N - 1 puppies in total.
  • n₁ - 1 black puppies
  • n₂ white puppies.

Now for the second selection, event B, the probability of selecting a white puppy is:

P = n₂/(N - 1)

Notice that because there are fewer black puppies than before (because we took one) the denominator decreases, thus, the probability of getting a white puppy increases.

So clearly event B depends on event A.

Now the question is:

Is A dependent on B?

No, A can not depend on event B, because event A happens first. The probability of the first event is always q = n₁/N.

If you want to learn more, you can read:

brainly.com/question/12138717

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2= 5.50 4= 10.50 here is your answer 
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What is the value of k?​
ELEN [110]

Answer:

10°

Step-by-step explanation:

According remote interior angles, 4k + 5 + 6k + 10 = 115.

10k + 15 = 115

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Hope this helps :)

Have a nice day!

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Answer:

y=-1/3x+4/3

Step-by-step explanation:

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A large but sparsely populated county has two small hospitals, one at the south end of the county and one at the north end. The
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Answer:

Step-by-step explanation:

Probability distribution of this type is a vicariate distribution because it specifies two random variable X and Y. We represent the probability X takes x and Y takes y by

f(x,y) = P((X=x,Y=y) ,

and given that the random variables are independent the joint pmf isbgiven by:

f(x,y) = fX(x) .fY(y) and this gives the table (see attachment)

(b) the required probability is given by considering X=0,1 and Y =0,1

f(x,y) = sum{(P(X ≤ 1, Y ≤ 1) }

= [0.1 +0.2 ] × [0.1 + 0.3]

= 0.3 × 0.4

= 0.12

Now we find

fX(x) = sum{[f(x.y)]} for y =0, 1 and similarly for fY(y)= sum{[f(x,y)]} for x=0, 1

fX(x) = sum{[f(x,y)]}= 0.1+0.3

= 0.4

fY(y) = sum{[f(x.y)]} = 0.1 + 0.2

= 0.3

Since fY(y) × fX(x) = 0.4 × 0.3

= 0.12

Hence f(x,y) = fX(x) .fY(y), the events are independent

(c) the required event is give by: [P(X<=1, PY<=1)]

= P(X<=1) . P(Y<=1)

= [sum{[f(x.y]} over Y=0,1] × [sum{[f(x.y)]}, over X= 0, 1

= 0.4 × 0.3

= 0.12

(d) the required event is given by the idea that: either The South has no bed and the North has or the the North has no bed and the South has. Let this event be A

A =sum[(P(X = 1<= x<= 4, Y =0)] + sum[P(X =0 Y= 1 <= x <= 3)]

A = [0.02 + 0.03 + 0.02 + 0.02] + [0.03 + 0.04 + 0.02]

A = [0.09] + [ 0.09]

A = 0.18

Pls note the sum of the vertical column X=2 is 0.3

3 0
4 years ago
afamily had dinner in a restaurant. If the cost of the cdinner was Rs. 2500, how much did the family pay with 10% service charge
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Step-by-step explanation:

Find 10%:

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Find the 13% vat:

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Find the total amount he got charged:

325 + 250 = 575

Total amount he had to pay for dinner + charges:

2500 + 575 = 3075

Answer: Rs. 3075

3 0
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