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Melody was mistaken. I have attempted all the questions... but I got them all incorrect. (3 was corrected)

1) What is the probability that, in a roll of 3 fair dice, the same number is not rolled twice?

I listed out all the posssible combinations that could have happen with 1. I got 20. I multiplied that by 6 and got 120. Then I know that 6 cubed is 216... so the I thought the probability must be 120/216 = 15/27.

2) What is the probability that, in a flip of 5 fair coins, there are at most 3 heads?

With 1 heads, that is 5!/4! outcomes = 5. With two and 3 heads... the outcomes are 5!/3!2! = 10. 10 + 10 + 5 = 25. 2 to the fifth power is 32. So the answer should be 25/32

3) What is the probability that a randomly chosen positive integer is divisible by 3 and 5, or 2?

The least common multiple of 2, 3, 5 is 30. There are 15 multiples of 2 and 10 multiples of 3, and 6 multiples of 5. 15 + 10 + 6 = 31. There are 2 multiples of 3*5, 3 multiples of 5*2, and 5 multiples of 3*2. 31 - 2 - 3 - 5 = 21 + 1 for counting 30. 22/30 = 11/15.

4) What is the probability that in 3 flips of a coin and 4 rolls of a die, there will be exactly 2 heads, and at least 2 2s?

I found of the probability of 2 heads was 3/8 through casework. For 2 2's I found the probabability was 1/216 also through casework

3/8 * 1/216 = 3/1728 = 1/576

5) If there is a 1/4 chance that I will win a game on every turn, what is the probability that the game takes at least 5 turns?

There was a 3/4 chance that it wasn't the winning turn... 3/4 * 3/4 * 3/4 * 3/4 * 1/4 = 81/1024.

6) There are 4 choices of ice-cream at a stand. There are 4 kids that each hate 1 flavor of ice-cream. All 4 flavors of ice-cream are hated by 1 kid each. In how many ways can all 4 kids receive a flavor of ice-cream that they do not hate, given that all 4 ice-cream flavors will be given?

I called the children A, B, C, and D. I called the ice cream flavors a, b, c, and d. I said that A hates a and B hate b and so on... A can receive it 3 ways, B can receive in 3 ways, etc. 3 choose 1 multiplied by 4... 12.

Now before you make any assumptions and go ahead deleting my problem, please see this.

Feb 23, 2020

#1
+108679
+1

Thank you for showing us what you have attempted.

That is what you should do EVERY time.

I for one cannot think straight with all those questions bombarding me.

One post is for one question.  I must have said that 1000 times.

But I will take a look at question 1

1) What is the probability that, in a roll of 3 fair dice, the same number is not rolled twice?

I listed out all the posssible combinations that could have happen with 1. I got 20. I multiplied that by 6 and got 120. Then I know that 6 cubed is 216... so the I thought the probability must be 120/216 = 15/27.

You listed all the possibilities .....  Where is your list?  Do you mean a list where one only happens once?

one of your combinations would be 1,2,3,

Now  if you do the same thing for the number 2 and add them together then this combination will appear again.

Hence I expect you have double counted.

See I can teach you if you show me where you are at!  Well maybe anyway.

With 3 rolls there is a 6^3 = 216 possible combination,

111       1 way

112        3ways

113         3ways

114         3ways

115         3ways

116         3ways

So that is 16 ways     (maybe that is what you did and one of us counted wrong?)

the same would  be for the one 5 numbers so

16*6= 96

so that appears to be   96/216 = 4/9

Edit:

This is the probability that a number will repeat.

The probability that a number will not repeat is 1-4/9 = 5/9

Now the idea is that if possible you see if my answer is correct.

If the answer is wrong then you politely and clearly explain why you think this.

(eg I put the answer in the answer box and the computer said it was wrong)

If the answer is correct then you try to see what I have done right and what you did wrong.

If you have questions then you ask for further explanation.

After you have digested all this and learned all that you can learn,

Then

You attempt the next question again by yourself and if you can not do it you present the next question and what you have tried in a completely new post.

-----------------------------------------------

Now digest all I have told you .

You could start by deleting questions 2 to 6 from this post.

One post = one question.

Feb 23, 2020
edited by Melody  Feb 23, 2020
#2
+29249
+5

For question 1 I would label the dice A, B and C.

Whatever number A shows, the probability that B shows something different is 5/6. The probability that C then shows a different number from both A and B is 4/6.  Hence overall probability is (5/6)*(4/6) = 20/36 = 5/9.

Feb 23, 2020
#4
+108679
0

Thanks Alan,

Alan's method is obviously the best way to so this problem.

Melody  Feb 23, 2020
#3
+1

Maybe, you could verify it this way:

The probability of having AT LEAST 2 of a kind =16 / 216. Since this applies to all 6 numbers, then the probability is:16 x 6 = 96 / 216. Now, this is the probability of at least 2 of a kind appearing. His question is:

"What is the probability that, in a roll of 3 fair dice, the same number is NOT rolled twice?". Then my understanding is that we want the opposite and subtract: [216 - 96] / 216 = 120 / 216 =5 / 9

Feb 23, 2020
#5
+108679
0

Yes that is right.

Melody  Feb 23, 2020