5LB Book - CHAP 22, #22 and CHAP 23, # 40

[anik1989]

I THINK THESE TWO QUESTION HAS TYPO, IF I AM WRONG PLEASE CLARIFY ME WITH EXPLANATION

1.(5LB CHAP 22, #22) In a class with 20 students, a test was administered, scored only in whole numbers from 0 to 10. At least one student got every possible score, and the average was 7.
Quanti ty A - 4

Quanti ty B- The lowest score that two students could have received

MY QUESTION IS HOW CAN A STUDENT GET every possible score??IS THIS LOGICAL?? SUPPOSE I SAY ONE OF MY FRIENDS IN MATH I SCORED 90, THEN ANOTHER I SAY IN MATH I SCORED 95. IS THIS LOGICAL??

2. CHAP 23, # 40
Jack has a cube with 6 sides numbered 1 through 6. He rolls the cube repeatedly until the first time that the sum of
all of his rolls is even, at which time he stops. (Note: it is possible to roll the cube just once.) What is the probability that Jack will need to roll the cube more than 2 times in order to get an even sum?
(A) 1/8
(B) 1/4
(C) 3/8
(D) 1/2
(E) 3/4

ANS (( Jack will only continue to roll the cube if the sum of the individual rolls is odd. For the first roll, this will
only occur if the number itself is odd; if Jack does not stop after the first roll, then, he must have rolled an odd
number for the first roll. For the second roll, in order for the sum of the first and second to be odd, Jack must now
roll an even (because odd + even = odd). You can rephrase the question: “What is the probability that Jack will roll an
odd first and an even second?” The probability of event A AND event B equals the probability of A times the
probability of B. Since the probability of odd = 1/2 and the probability of even = 1/2, the probability of the first
number being odd AND the second number being even is (1/2)(1/2) = 1/4.))

IS THIS NOT TYPO?? THE QUESTION SAYS WHEN SUM OF ALL HIS ROLL IS EVEN THEN HE STOPS. IF HE ROLLS THIRD TIME A EVEN THEN SUM WILL BE ODD. BUT AUTHOR ONLY CALCULATES “What is the probability that Jack will roll an
odd first and an even second?”

[tommywallach]

Anik,

FIRST of all, please STOP titling all your posts as typo when the issue seems to be that you don't understand the question. There is no typo here, and it's a little frustrating that you keep expecting a typo because you don't understand the math. Please don't make me change all your titles (I've changed all the ones you've posted up to now). Assume you don't understand the question, not that the book is wrong.

SECOND, please only post ONE question per thread, and title that thread with the Book name, chapter number, and question number at issue, not some random words.

THIRD, please no caps, and try to format your questions and the questions you post with correct grammar and spacing, as per the original version.

OKAY: now to your questions.

To your first point, the sentence says that at least one student got every score. This means that of the 11 possible scores (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10), at least one student had to get each score. Your reading on it is patently illogical, so of course it's clear what is meant. This is not an error.

For the second question, the only thing we need to know is what the odds are that Chip gets to a third roll. We don't need to calculate what happens on that roll, because the question is only about whether he needs to roll more than twice. At this stage (after two rolls where the sum is still odd), Chip now MUST roll more than two times.

-t

[anik1989]

thsnk you sir, i wont make any title further

[anik1989]

for the second one , the explanation strats like this
"Jack will only continue to roll the cube if the sum of the individual rolls is odd. For the first roll, this will only occur if the number itself is odd; if Jack does not stop after the first roll, then, he must have rolled an odd number for the first roll. For the second roll, in order for the sum of the first and second to be odd, Jack must now
roll an even (because odd + even = odd)..."

SINCE , Jack will only continue to roll the cube if the sum of the individual rolls is odd.SO, he must try third times whether it odd or even. if it is even then odd+even+even=odd ,then he will try for 4th time to make the sum of the individual role even..and so on..

so , please tell me where is my lacking of understanding?? if you are not busy would you please do this math for me?? because solution is better than conversation

[tommywallach]

Anik,

The problem you're having has nothing to do with the math, but with the language of the question.

Think of it this way:

Tommy has a 1/2 chance of setting the house on fire every time he tries. What are the odds it takes him more than two tries?

Well, the first time, I have a 1/2 chance of FAILING to light the house on fire. If I succeed the first time, then I won't NEED to try a second time, so we NEED to know that I fail the first time. Now, to get to a THIRD try, I need to fail on the first AND second tries (otherwise, I'd be finished). So the odds that I will require more than two tries are the odds that I FAIL on the first AND second tries (in this case, 1/2 * 1/2 = 1/4). I don't need to CALCULATE the third try, because the only question was whether I will REQUIRE that third try.

Make more sense?

-t

[anik1989]

Anik,

The problem you're having has nothing to do with the math, but with the language of the question.

Think of it this way:

Tommy has a 1/2 chance of setting the house on fire every time he trees. What are the odds it takes him more than two tries?

Well, the first time, I have a 1/2 chance of FAILING to light the house on fire. If I succeed the first time, then I won't NEED to try a second time, so we NEED to know that I fail the first time. Now, to get to a THIRD try, I need to fail on the first AND second tries (otherwise, I'd be finished). So the odds that I will require more than two tries are the odds that I FAIL on the first AND second tries (in this case, 1/2 * 1/2 = 1/4). I don't need to CALCULATE the third try, because the only question was whether I will REQUIRE that third try.

Make more sense?

-t

I understand now. thank you for your example, sir

[tommywallach]

Glad to help!

-t