Question

A program to simulate coin tosses is used to simultaneously toss 1,000 coins. If the number of heads displayed in 5 trials of the simulation total 528, 486, 492, 545, and 505, what is the experimental probability of getting heads on a coin toss? Round your answer to the nearest percent.

50%
51%
49%
53%

Answer 1

Is there anyone that's good a prob. and stats that can help me?

Answer 2

Thanks, I already answered the question but I didn't get it correct. Do you think you can help me with 5 problems, John?

Answer 3

I can attempt, yes.

Answer 4

Omg thank you! Hold on a second and I will type them :D

Answer 5

A die of unknown bias is rolled 20 times, and the number 3 comes up 6 times. In the next three rounds (the die is rolled 20 times in each round), the number 3 comes up 6 times, then 5 times, and finally 7 times. Which experimental probability is consistent with this simulation?

Answer 6

The probability of 3 is 15%
The probability of 3 is 30%
The probability of 3 is 45%
The probability of 3 is 60%

Answer 7

You will probably need the average of the numbers it was rolled. So
\[\frac{ (6+6+5+7) }{ 4 } = \frac{ 24 }{ 4 } = 6\]

Answer 8

6 is what percent of 20?

Answer 9

30?

Answer 10

\[6 = \frac{ x }{ 100 } \times 20\]

Answer 11

Correct so the probability of attaining a 3 should be 30%

Answer 12

Thank you! Okay! Next question..
A person playing a game of chance has a 0.25 probability of winning. If the person plays the game 20 times and wins half of that number of times, what is the difference between the theoretical probability and the experimental probability of that person winning ?

Answer 13

0.50
0.25
0.30
0.75

Answer 14

You know the probability of winning is 0.25

When the person played 20 times they won half the time so,\[\frac{ 10 }{ 20 } = \frac{ 1 }{ 2 }= 0.50\]

Now we are asked to find the "difference" between the theoretical(0.25) and the experimental (0.50).

Answer 15

Difference refers to subtraction of one term by another and in this case we know both terms.

Answer 16

So it's .50?

Answer 17

Nope. We need to subtract one term from the other.

Answer 18

Okay so we subtract .50-.25? that equals .25

Answer 19

Correct!

Answer 20

Thanks! Okay next question :D

A simulated game of chess is programmed between two computers. The game is supposed to be biased in favor of player B winning 4 out of 5 times. Which is the most suspicious set of outcomes of 30 games played between the two computers?

Answer 21

AABABBABBBBBBBBABBBABBBBBBBBA
ABBABBAABBBBBABBBBBBBBBBBBBABB
ABBABABABABBABABABABABABABAA
ABBABBBABBBBBBBBABBBA

Answer 22

B should win 4/5 times or,\[\frac{ 4 }{ 5 } = 0.80\]

Answer 23

a. AABABBBABBBBBBBBABBBABBBBBBBBA

b. ABBABBAABBBBBABBBBBBBBBBBBBABB

c. ABBABABABABBABABABABABABABABAA

d. ABBABBBABBBBBBBBABBBABBBBBBBBB

^sorry, these are the correct choices

Answer 24

If you add up the total number of times B won and divide it by the 30 games played. It should be 4/5 or 0.80.

Answer 25

So it's B? Are you sure?

Answer 26

I am not sure, i have not performed the operation.

Answer 27

I looked the question up and the person said it's not b. It's something else.

Answer 28

It is not B, B wins 24 times out of 30 or, \[\frac{ 24 }{ 30 } = \frac{ 12 }{ 15 } = \frac{ 4 }{ 5 } = 0.80\]

Answer 29

Did not say it was B, I was referring to "player B" before.

Answer 30

Oh, I thought you were saying it was B XD. Okay, so B won four out of 5 times...and you divide it by 30? O.o

Answer 31

No B won 24/30 times and i just simplified, hence the = signs above.

Answer 32

24/30 = 0.80 in your calculator

Answer 33

Okay, I got .80 on my calculator.

Answer 34

So it is correct and does not look suspicious.

Answer 35

You are looking for the "most" suspicious which means the one that is way different from your 4/5 or 0.80

Answer 36

I don't think it looks suspicious.

Answer 37

It doesn't at all. It is exactly 0.80 so it is not our suspicious game.

Answer 38

Okay. So how do we find the one that's the most suspicious?

Answer 39

Find the number of times B won in each game and divide it by the total number of games played.

Answer 40

a. AABABBBABBBBBBBBABBBABBBBBBBBA, A= 7 , B = 23

b. ABBABBAABBBBBABBBBBBBBBBBBBABB, A = 6, B = 24

c. ABBABABABABBABABABABABABABABAA, A= 15, B =15

d. ABBABBBABBBBBBBBABBBABBBBBBBBB, A = 5 , B = 25

Answer 41

Now which looks most suspicious?

Answer 42

Okay, so A) B won 22 times? C)14 and D) 26 times?

Answer 43

I got different values than you did?

Answer 44

We both are close though even with both our values which of them looks suspicious? And where B does not win 4/5 times

Answer 45

B meaning Player B

Answer 46

C looks suspicious to me.

I probably didn't count the B's correctly.

Answer 47

It is alright.
Correct because in my count B won half the time not close to 4/5.
And in your count B won almost half the time which is also not close to 4/5.

Answer 48

So it is C? :D

Answer 49

Correct!

Answer 50

C is the most suspicious game with 0.50 probability of winning which is farther from 0.80 compared to the other games.

Answer 51

Thank you! :D okay, just 2 more questions!

A sharpshooter in a training simulation has a miss rate of 1%. If H represents a hit and M represents a miss, which is the most likely valid result of a simulation model of 30 practice shots?

Answer 52

.HHHHHHHHHHHHHHHHHHHHHHHHHHHHHH

b. MHHHHHMHHHHHHHHHHHHHMHHHHHHHHH

c.HHHHHHHMHHHHHMHHHHHHMHHHHHHMHH

d.HMHHHHHHHHHHHHHHHHHHHHMHHHHHHH

Answer 53

*a

Answer 54

Most likely "valid" result so we are looking for the results that most depicts a 1% miss rate.

Answer 55

Looks like A to me....

Answer 56

First we should ask:
What is 1% of 30 games?

Answer 57

.30?

Answer 58

Correct it is less than 1 so we are looking for when he is least likely to miss.

Answer 59

Okay :D so do we have to count the H's or M's?

Answer 60

You are correct all of other games have a higher miss rate than the game A which has a miss rate of 0% closest to 1%

Answer 61

I'm pretty sure it's a....

Answer 62

That is correct.

Answer 63

You are correct all of other games have a higher miss rate than the game A which has a miss rate of 0% closest to 1%

Answer 64

Thanks! One more :D A soccer team wins 65% of its matches, and 15% of its matches end in a draw. If the team is scheduled to play 20 matches, about how many matches is it expected to lose?

Answer 65

Well you know that A soccer team will win 65% and 15% will be draws so out of 100% the rest must be loses.