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# Incredibly Difficult Probability Questions!!!

0
301
5
+14

1. Jenn randomly arranges the three digits 1, 2, and 3 to create a three-digit number. What is the probability that her number is a multiple of 4?

2. Rolling four standard dice, what is the probability that the product of the four results is even?

3. Four slips of paper labeled A, B, C, and D are drawn from a hat at random, one at a time, without replacement. What is the probability of drawing C first and D last? Express your answer as a common fraction.

4. A bag contains 3 red balls, 4 green balls, and 5 yellow balls. If balls are drawn one at a time without replacement, what is the probability that the first yellow ball is drawn on the eighth draw? Express your answer as a common fraction.

5.  The Village League uses a best-two-out-of-three series to crown its champion. If the probability of Team Alpha beating Team Beta is 60% for every game, what is the probability that Beta wins the championship? Express your answer as a common fraction.

6. A spinner is divided into 5 sectors as shown. Each of the central angles of the sectors 1 through 3 measures 60% while each of the central angles of sectors 4 and 5 measures 90%. If the spinner is spun twice, what is the probability that at least one spin lands on an even number? Express your answer as a common fraction.

7. If a committee of six students is chosen at random from a group of six boys and four girls, what is the probability that the committee contains the same number of boys and girls? Express your answer as a common fraction.

8. In 2010, the probability of Mark sinking a free throw was twice what it was in 2009, and yet the probability of him sinking exactly two out of three free throws was the same in both years. What was the probability of Mark sinking a free throw in 2010, given that the probability was greater than zero?

May 12, 2019

#1
+14
-1

All I really need is number 6 and 5! Thanks!

May 12, 2019
#5
+107115
0

Well do not post all the others then.

When I see a long list of questions like this I rarely will read or answer them.

If I am feeling excessively annoyed then I delete parts or hide the whole question.

We are not here to do all of a person's homework for them.

And that is what is looks like if multiple questions are being asked.

Melody  May 13, 2019
#2
+4330
+1

1. 3!=6 ways to arrange the three-digit number...and 132, 312 are multiples of 4. Thus, 2/6=1/3?

May 12, 2019
#3
+106539
+2

6. A spinner is divided into 5 sectors as shown. Each of the central angles of the sectors 1 through 3 measures 60° while each of the central angles of sectors 4 and 5 measures 90°. If the spinner is spun twice, what is the probability that at least one spin lands on an even number? Express your answer as a common fraction.

P(even on one at least of the spins) =   1 - P(odd on both spins)

P(odd on 1st spin)  =  [60 + 60 + 90] / 360  =  210/360  = 7/12

And we have the same probability of an odd on the second spin

So

P(odd on both spins)  =  7/12 * 7/12  =   49/144

So....P (even on at least one of the spins)  =  1 - 49/144  =  95 / 144

May 12, 2019
#4
+569
+1

Most probability problems are easier to solve when solving for the chance something won't happen, for some reason, although trying to do that with 5 appears to give me -8%, so I'm going to do the longer route.

We can start by first listing all ways that Beta can win. Namely, B-B-A, B-B-B, B-A-B, and A-B-B. BAB and ABB are technically the same thing because the order doesn't affect the outcome, so I will solve them once and double the chance. Similarly, the final game has no effect on the outcome if Beta wins the first 2 games, so we can simplify BBA + BBB to BB_ (the blank is there to represent the unplayed final game).

BB: The chance of Beta winning the first 2 games is equal to their chance of winning one game squared (chance of B winning * chance of B winning = chance of B winning in any 2 given times, a.k.a. the first 2 times now.) 40% * 40% = 16%, so BB = 16%.

ABB and BAB: We can assume Alpha won the first game, as the order doesn't matter (it's just who won more games, not who won which game when). Therefore, ABB = 60% * 40% * 40% = 9.6%.

Now that we have all of the outcomes covered, we can add them. BB + BAB + ABB = 16% + 9.6% + 9.6% = 35.2% chance that team Beta wins the championship.

Hope this helped!

May 13, 2019