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Help in probability... here are some questions..


Curly213

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1. If Roger lives at point A and works at point B, how many different ways can he travel to work if he must always take the shortest path?

2. How many ways can Mike and John sit down at the table so that they don’t sit directly across from each other? 224

3. How many different ways can you arrange 5 beads on a metal ring? 24

4. A parliamentary committee comprising of six members is to be formed from a group of 6 ruling party members and 5 opposition members. In how many ways can this be done if the committee comprises exactly three ruling party members? 200

5. If six men and six women form a group of 4 people, how many different groups can be created that have at least one man? 480

6. If any number from set A is multiplied by any number from set B, what is the probability that the product is a multiple of 4? 2/5

A = {21, 22, 23, 24, 25}

B = {23, 24, 25, 26, 27}

7. A jar contains 6 red balls and 4 black balls. What is the probability of getting two red balls? What is the probability of getting 2 red balls and 2 black balls? 1/3 and 3/7

8. A coin has been weighted so that it is 6 times as likely to come up heads as it is tails. If the coin is flipped, what is the probability that the coin comes up tails? 1/7

9. A bowl contains 10 apples, 2 of which are bad. If someone randomly selected four apples from the bowl, what is the probability that at least one of the apples is bad? 2/3

10. If five coins are tossed, what is the probability that at least three of the coins come up heads? 1/2

 

 

For the first two ex are pics but i cant paste them into here... have to find out how it works ..would be nice if someone could help me with this matter...

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9. A bowl contains 10 apples, 2 of which are bad. If someone randomly selected four apples from the bowl, what is the probability that at least one of the apples is bad?

 

My answer: 2/3

 

In this situation it's easiest to first find the probability of selecting all good apples, then subtracting that from one in order to find the probability of at least selecting one bad apple.

P(all good)=

(8/10)*(7/9)*(6/8)*(5/7)=

(4/5)*(7/9)*(3/4)*(5/7)= (after you cancel tops and bottoms)

(1)*(1/3)*(1)*(1)= 1/3

P(all good)= 1/3

 

P(at least one bad)= 1-P(all good)

P(at least one bad)= 1-(1/3)

P(at least one bad)= 2/3

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7. A jar contains 6 red balls and 4 black balls. What is the probability of getting two red balls? What is the probability of getting 2 red balls and 2 black balls? 1/3 and 3/7

..

 

Note: This question is anbiguous as it does nto describe properly the drawing of the balls. We actually want to know whether the selection simultaneous. I will just assume that the balls are chosen one at a time.

 

Probability questions can be solved by laying out the information carefully and asking the key questions for this question type.

 

For the second part:

 

we want RRBB (ignoring order)

 

so we can lay out the question __(R1) __(R2) __(B1) __(B2)

 

Then we place the probability of getting the result we want for each slot in that slot.

 

6/10 (R1) 5/9 (R2) 4/8 (B1) 3/7 (B2)

 

So, to find the probability of getting RRBB, we multiply these numbers.

 

Note: You should wait to do operations until the end of the question or as long as possible.

 

Now we need to consider order, as long as the balls selected have two red and two black we have met the conditions of the question. Therefore, we want to make sure to count all possible rearragements of the selction order that would give this result.

 

The forumla to find rearragements is

(total things rearranged)!/(# thing 1)!(number of thig 2)! ...

 

In this case, that would give us:

4!/(2!)(2!) = 6

 

so our answer should be:

6/10 (R1) 5/9 (R2) 4/8 (B1) 3/7 (B2) * 6 = 3/7

 

I hope this helps.

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