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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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(4) we need 3 out of 6 (ruling), and 3 out of 5 (oppsition), right ?

6C3 * 5C3 = 20 * 10 = 200

 

(5) total possibilities = 12C4 , cases in which only women = 6C4

so required no = 12C4 - 6C4 = 495-15 = 480

 

(6) total cases = setA*setB = 5*5 = 25, right ?

we'll have multiple of 4 when

"24" from set A is multiplied with any of the 5 values from set B ------> 5 cases

and

"24" from set B is multiplied with any of the 5 values from set A ------> 5 cases

so favorable cases = 10/25 = 2/5

 

(7) question should read "without replacement". then, prob(1st red)= no of red/total=6/10, next red ball has prob=5/9(bcos 9 are left)

so (6/10)*(5/9) = 1/3

 

for 2 red/2 black, im getting 1/14 (i'll chaeck later wether im wrong or the ans given here is)

 

 

(8) "6 times as likely" means odds, so if odds are 6 to 1, then ratio of heads:tails = 6:1, so probability(tails) = 1/(6+1)

 

(9) 8 are good, so prob(at least one bad) = 1 - (prob of no bad apples) = 1 - [(8*7*6*5)/(10*9*8*7)] = 2/3

 

(10) at least 3 heads = 3 heads or 4 heads or 5 heads = 10/32 + 5/32 + 1/32 = 16/32 = 1/2 [explanation is long, i'll explain in case u dont understand]

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3. The arrangements are taking place in a circle. In a circle, some arrangements are the same just by rotating the circle so you need to take those off. Set a point of reference, and permutate the rest. You have 1 bead fixed, and arrange the rest. So 4! ways.

 

4. 6c3*5c3 [of the 6 ruling party ones choose 3, and of the opposing 5 choose 3]

 

5. If you don't want to break your head thinking about the possible ways men can have at least 1, you can just find the ways in which it is not possible for a male to be chosen and subtract from the total possible ways you can choose 4 from a group of 12. 12c4-6c4

 

6. I don't have an easy way for this, you'll have to lay out the possible outcomes. I'll love to know the fast way.

 

7. Two red balls [the pr. of the ball being red on the first pick is 6/10, assuming you don't replace the balls, then the pr of getting a second red is 5/9. 6/10*5/9=30/90=1/3] You can figure the next one with the same reasoning.

 

8. The set P heads + P tails must equal 1.

P heads = 6 p tails.

plug in 6 p tails for p heads in the first equation. you get 6 ptails + ptails=1

7ptails=1 or ptails=1/7 [just remember all probabilities must add to one]

 

9. Again, you can lay out all the possible ways you can select 4 apples and subtract all the possible ways you can get only good apples.

 

(10c4-8c4)/10c4

 

10. 1- pr[less than 3 heads]

pr less than 3 heads = (5c2) pr(heads)^2*pr(tails)^3

+ (5c1) pr(heads) * pr(tails)^4

+(5c0) pr (tails)^4

{Just added} [To get 2 heads you need HHTTT, how many ways can you choose 2 heads from 5? 5c2 and the probabilities of each occurring (since this are independent) heads*head*tails*tails*tails=heads^2*tails^3. You can follow this same reasoning for the other probabilities]

Do those add them up and subtract from 1.

 

There may be some minor errors...=p

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Let's try this!

The first can be either black or red so lets pick red. P[red]=6/10

the second can also be red or black. Let's pick red again. P[red]=5/9

The other 2 must be black. So 4/8*3/7

Case 1.

Its probability is 6/10*5/9*4/8*3/7 =1/14

 

Let's choose black first.

4/10 then another black 3/9 and the last 2 must be red.

4/10*3/9*6/8*5/7=1/14

Case 2. also =1/14. Do you see a cycle here. All the numbers in the numerator are the same and all the numbers in the denominator are the same; just arranged differently.

So lets do something different. 4c2*1/14 = 3/7 (4c2 just to find the number of ways to have 2 reds and 2 blacks)

 

There are going to be 6 cases each with probability 1/14. So we need to add all those small pieces.

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For the 10th question we can use Binomial Distribution formula

 

If the probability of A happening is P, then the probability of A happening exactly K times, if repeated N times is given by

 

NcK * p ^ k * (1 - p) ^ (n - k)

 

In this case

 

required = 1 - [0 heads + 1 head + 2 heads]

 

= 1 - [5c0 * (1/2)^0 * (1/2)^5 +

5c1 * (1/2) ^ 1 * (1/2)^4 +

5c2 * (1/2) ^ 2 * (1/2)^3 ]

 

= 1 - 16/32

 

= 1/2

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