Probability Problems

[Gasedo]

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Is somebody willing to explain these problems?
I don’t know if there are a lot of probability problems on the DAT, but hopefully there are going to be reasonable. 🙁

A drawer contains 2 blue, 4 red and 2 yellow socks. If to socks are to be randomly selected from the drawer, what is the probability that they will be the same color?
A 2/7
B 2/5
C 3/7
D 1/2
E 3/5

A fair coin is to be tossed 8 times. What is the probability that more of the tosses will result in heads that will result in tails?
A 1/4
B 1/3
C 87/256
D 23/64
E 93/256

 

[lafootdoc]

yo, my advice to you is to get the barrons dat book. Its only like 20 bucks and i swear it teaches the math probobilty stuff really really good

Is somebody willing to explain these problems?
I don’t know if there are a lot of probability problems on the DAT, but hopefully there are going to be reasonable. 🙁

A drawer contains 2 blue, 4 red and 2 yellow socks. If to socks are to be randomly selected from the drawer, what is the probability that they will be the same color?
A 2/7
B 2/5
C 3/7
D 1/2
E 3/5

A fair coin is to be tossed 8 times. What is the probability that more of the tosses will result in heads that will result in tails?
A 1/4
B 1/3
C 87/256
D 23/64
E 93/256

 

[howui3]

A fair coin is to be tossed 8 times. What is the probability that more of the tosses will result in heads that will result in tails?
A 1/4
B 1/3
C 87/256
D 23/64
E 93/256

Here is a similar question and its solution:

What is the probability of six tails out of nine tosses of a fair coin?

p = probability of getting a tails (1/2)
C(x,y) where x = 9 and y = 6.
C(x,y) = x!/[y!*(x-y)!]

The equation to solve this is (just memorize it):

C(x,y)*p^(x)*(1-p)^(x-y)

Plug in the values for x and y and you should get 21/128

Also check out this website:

http://www.saliu.com/theory-of-probability.html

Good luck.

 

[Gasedo]

Here is a similar question and its solution:

What is the probability of six tails out of nine tosses of a fair coin?

p = probability of getting a tails (1/2)
C(x,y) where x = 9 and y = 6.
C(x,y) = x!/[y!*(x-y)!]

The equation to solve this is (just memorize it):

C(x,y)*p^(x)*(1-p)^(x-y)

Plug in the values for x and y and you should get 21/128

Also check out this website:

http://www.saliu.com/theory-of-probability.html

Good luck.

Thanks, I really apreciate it.

 

[howui3]

Thanks, I really apreciate it.

came across some more probability question:

1) How many ways to select a panel of 3 people from a group of 5 people?

solution: nCk = n!/k!(n-k)! and in our case 5C3=5!/3!(5-3)! = 5!/3!2! = (5*4*3*2*1)/(3*2*1*2*1) = 10



2) How many ways can the letters of the word AGREE be arranged?

solution: since there are 2 E's = 5!/2! = 60



3) a fair 6-sided die is rolled 5 times. What is the probability that on exactly 3 rolls the number of dots on the die is no greater than 4?

solution: First remember that there are 6 possible outcomes.

Probability of getting 4 or less (1,2,3,4) = 4/6 or 2/3.

Probability of getting 5 or more (5,6) = 2/6 or 1/3.

Probabilty of just getting the wanted number on the die = (2/3)(2/3)(2/3)(1/3)(1/3)=8/243

Now we want 3 events in 5 rolls, must use the compination equation: 5C3=5!/3!2! = 10

So TOTAL Probability= 8/243 * 10 = 80/243

 

[Gasedo]

came across some more probability question:

1) How many ways to select a panel of 3 people from a group of 5 people?

solution: nCk = n!/k!(n-k)! and in our case 5C3=5!/3!(5-3)! = 5!/3!2! = (5*4*3*2*1)/(3*2*1*2*1) = 10



2) How many ways can the letters of the word AGREE be arranged?

solution: since there are 2 E's = 5!/2! = 60



3) a fair 6-sided die is rolled 5 times. What is the probability that on exactly 3 rolls the number of dots on the die is no greater than 4?

solution: First remember that there are 6 possible outcomes.

Probability of getting 4 or less (1,2,3,4) = 4/6 or 2/3.

Probability of getting 5 or more (5,6) = 2/6 or 1/3.

Probabilty of just getting the wanted number on the die = (2/3)(2/3)(2/3)(1/3)(1/3)=8/243

Now we want 3 events in 5 rolls, must use the compination equation: 5C3=5!/3!2! = 10

So TOTAL Probability= 8/243 * 10 = 80/243

Hopefully I am not going to get a problem like the 3rd one. It takes time to do it and under that pressure you can mess up easily. Do you remember if you had many probabilities problems on the test?
Some people were saying they got one or two. 🙁

 

[IndDent]

Get REA Algebra & Trig book, $25 at barnes. Or at least check out the probability section of the book. It is amazing, and makes it really really simple to understand. Not only that, it has 100s of practice problems.

I was absolutely lousy at this section of the DAT, but pulled off a 23 on the real thing, and honestly it's all thanks to this REA book for the Trig and probability section. Rest, I relied on the Kaplan blue dat book formulas that I memorized.

Take it easy,
Indy

 

[jessica22]

Howui,

How did you get 21/128? I plugged it in and didn't get it..... 😕

Here is a similar question and its solution:

What is the probability of six tails out of nine tosses of a fair coin?

p = probability of getting a tails (1/2)
C(x,y) where x = 9 and y = 6.
C(x,y) = x!/[y!*(x-y)!]

The equation to solve this is (just memorize it):

C(x,y)*p^(x)*(1-p)^(x-y)

Plug in the values for x and y and you should get 21/128

Also check out this website:

http://www.saliu.com/theory-of-probability.html

Good luck.

 

[americanpierg]

C(x,y) = x!/[y!*(x-y)!]

This equation is such a life saver, randomly probability questions about microstates in entropy for gchem poped up on an exam and asked about the prob of having a certain number of molecules on a certain side of a certain rectugular box, memmorize that equation and u will be golden

 

[americanpierg]

Howui,

How did you get 21/128? I plugged it in and didn't get it..... 😕

(9*8*7*6*5*4*3*2*1) / [6*5*4*3*2*1 * (3*2*1)] = C(x,y) = 84

84* (1/2)^9 * (1-1/2)^(9-6) = 21/1024

howui got 21/128 because instead of C(x,y)*p^(x)*(1-p)^(x-y), he had C(x,y)*p^👍*(1-p)^(x-y), he took p to the y power instead p to the x

 

[BKN]

(9*8*7*6*5*4*3*2*1) / [6*5*4*3*2*1 * (3*2*1)] = C(x,y) = 84

84* (1/2)^9 * (1-1/2)^(9-6) = 21/1024

howui got 21/128 because instead of C(x,y)*p^(x)*(1-p)^(x-y), he had C(x,y)*p^👍*(1-p)^(x-y), he took p to the y power instead p to the x

Actually, I think Howui stated the formula incorrectly,but got the answer right. You can keep track of this easier if your use the other nomenclature:

n = number or trials
K= number of successes
n-k= number of failures
p = probability of success
1-p = q = probability of failure

In this case p = q =0.5

So the probability of k successes in n trials if the probability of success in 1 trial is p = (n!/k!(n-k!)) p^k*q^n-k

In this case 84*0.5^6*0.5^3 = 84*0.5^9 = 84/512 = 21/128

 

[jessica22]

Thank you so much you guys. You guys are so nice.

 

[howui3]

Actually, I think Howui stated the formula incorrectly,but got the answer right. You can keep track of this easier if your use the other nomenclature:

n = number or trials
K= number of successes
n-k= number of failures
p = probability of success
1-p = q = probability of failure

In this case p = q =0.5

So the probability of k successes in n trials if the probability of success in 1 trial is p = (n!/k!(n-k!)) p^k*q^n-k

In this case 84*0.5^6*0.5^3 = 84*0.5^9 = 84/512 = 21/128

Thank you for the correction. It has been a while since my DAT.

Good luck!