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You are a tossing a balanced die that has probability 1/6 of coming up 1 on each toss. Tosses are independent. You’re interested in how long you must wait to get the first

a.The probability of a 1 on the first toss is 1/6. What is the probability that the first toss is not a 1 and the second toss is a 1?

b.What is the probability that the first two tosses are not 1s and the third toss is a 1?

c.See a pattern??? What is the probability that the first 1 occurs on the fourth toss? On the fifth toss? Give the general result: What is the probability that the first 1 occurs on the kth toss?

2006-09-20 18:49:34 · 7 answers · asked by andy 1 in Science & Mathematics Mathematics

7 answers

a.
Probability of tossing non-1 on first roll: 5/6
Probablity of tossing 1 on second roll: 1/6
5/6 * 1/6 = 5/36 (13.89%)

b.
Probability of tossing non-1 on first roll: 5/6
Probability of tossing non-1 on second roll: 5/6
Probability of tossing 1 on thrid roll: 1/6
5/6 * 5/6 * 1/6 = 25 / 216 (11.57%)

c.There is a probablity of 5/6 for every non-1 roll, and a probability of 1/6 for the last roll (for the 1). Therefore, the probility that the first 1 appears on the kth toss is
(1/6) * (5/6)^(k-1)

2006-09-20 21:02:33 · answer #1 · answered by Phu N 2 · 0 0

the three 6 sided cube contemporary 216 diverse mixtures. you will %. one quantity. you have 18 cube factors, and one quantity which will seem three times. the probabilities of your quantity coming up are subsequently 3/18, which may well be simplified to a million/6. each 6 rolls, your one quantity would desire to arise. i assume we can follow one quantity. i will %. one. Out of all the mixtures, a million looks as quickly as in 25 activities. that ought to furnish 50 factors. The form of circumstances that one looks two times is 10. that ought to furnish you 40 greater factors. One looks three times purely as quickly as, so because it is 6 factors. 50+40+6 is ninety six factors. you will purely get ninety six factors.

2016-12-18 14:07:16 · answer #2 · answered by cordell 4 · 0 0

(5/6)^(k-1) / 6 or,

5^(k-1) / 6^k

2006-09-20 19:04:54 · answer #3 · answered by Joe C 3 · 0 0

a) .138888888... percent chance
b) .1157
c).096,.803,and i think it is 5/6^n times of rolls before 1, and then time .1666, but i very well may be wrong lol

2006-09-20 18:56:18 · answer #4 · answered by David 5 · 0 0

Any beginners statistic book will have this in it. I am sure there are thousands of sited depicated toprobablility of dice rolls.

2006-09-20 18:51:57 · answer #5 · answered by DoctaB01 2 · 0 0

a. 5/6X1/6 = 5/(6^2)
b. 5/6X5/6X1/6 = (5^2)/(6^3)
c. 5^(k-1)/6^k

2006-09-20 19:03:46 · answer #6 · answered by Anonymous · 0 0

http://nces.ed.gov/nceskids/probability/
http://www.math.princeton.edu/matalive/Probability/ProbabilityLab1/Rolling.html
http://www.edcollins.com/backgammon/diceprob.htm

2006-09-20 18:54:05 · answer #7 · answered by yoursuperman000 2 · 0 0

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