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topjm [15]
3 years ago
5

One urn contains one blue ball (labeled b1) and three red balls (labeled r1, r2, and r3). a second urn contains two red balls (r

4 and r5) and two blue balls (b2 and b3). an experiment is performed in which one of the two urns is chosen at random and then two balls are randomly chosen from it, one after the other without replacement. what is the total number of outcomes of this experiment?
Mathematics
2 answers:
Rina8888 [55]3 years ago
5 0

Answer:

12 outcomes

Step-by-step explanation:

Given:-

Urn 1 : 1 blue ball & 3 red balls

Urn 2 : 2 blue & red balls.

Find:-

One of the two urns is chosen at random and then two balls are randomly chosen from it, one after the other without replacement. what is the total number of outcomes of this experiment?

Solution:-

- The number of possible outcomes when selecting one of the 2 urns available is = 2, since we can either choose Urn 1 or Urn 2.

- Once the urn is selected each urn has a total of 4 balls ( Red & Blue ). We are to choose 2 balls from the chosen urn. The number of combinations for selecting 2 out of 4 available is = 4C2 = 6 possibilities.

- Then the total number of combinations are:

                 Total outcomes = 2 * 6 = 12 outcomes

Flura [38]3 years ago
3 0

Answer:

12 possibilities

Step-by-step explanation:

In the first urn, we have 4 balls, and all of them are different, as they have different labels, so the group of two red balls r1 and r2 is different from the group of red balls r2 and r3.

The same thing occurs in the second urn, as all balls have different labels.

The problem is a combination problem (the group r1 and r2 is the same group r2 and r1).

For the first urn, we have a combination of 4 choose 2:

C(4,2) = 4!/2!*2! = 4*3*2/2*2 = 2*3 = 6 possibilities

For the second urn, we also have a combination of 4 choose 2, so 6 possibilities.

In total we have 6 + 6 = 12 possibilities.

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Find the sum of even numbers from 34 to 234 inclusive
valentinak56 [21]
We have 

AP = 34,36,38......,234

a, (first term) = 34

n ( total terms ) =?

d ( common difference ) = 2

so, a +(n-1)d =an

34+(n-1)*2 = 234

(n-1)2 = 234-34

n-1 = 200/2

n = 100+1

n =101

Now sum of the even no.

Sn = n/2( 2a+9n-1)d)
       
       = 101/2 (2*34 +100*2)
 
        = 101/2 * 268
 
          =101 *134
 
         = 13534
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3 years ago
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Alenkinab [10]
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3 years ago
How many solutions does this system of equations have?
Alja [10]

Answer:

Infinitely many

Step-by-step explanation:

The second equation is basically multiplying the first equation by 2, so this means that they will have the same rate of change and initial value. Therefore, there are infinitely many solutions.

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3 years ago
Select the correct answer
NARA [144]

I think the answer is $39

3 0
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A bathtub holes about to hectoliters of water if 25 L of water are let down the drain how many liters of water are still in the
DanielleElmas [232]
Answer:
75 liters are left in the bathtub

Explanation:
First, we should know that a hectoliter is a unit of capacity that is equal to 100 liters.

We are given that the bathtub holds 1 hectoliter of water.
This means that the bathtub holds 100 liters of water.
We are also given that 25 liters of water were drained from the bathtub.
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Hope this helps :)
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