The probability that the student will pass is; P(pass)
=
P(10) + P(9) + P(8) =
[10C10 + 10C9 + 10C8] / (.5)10 =
56/1024 ~
~ 0.0547 ~ 5.47%
where nCr =
n!/[r!(n-r)!]
It is 0.3438
The difference between "at least" and "at most" is not restricted to probability. The difference is simply one between the precise meaning of the phrases in every day English language.
A fair chance of winning is more than 50% chance of winning. Therefore probability = 0.5 We need to find a fair chance of winning atleast one match. 1-(5/6)^n > 0.5 hence, n=4 QED (quite easily done, :-p)
Why do you think it is a question? It is a homework assignment.
powers of 10 are like this 10(small 1) 10(small 2) exetra i need help on the same question but atleast this might help someone
Total of 30 questions, out of wich you have to answer atleast 21 correctly to get the permit.
7/128, or about 5.5% The student has a 1/2 probability of getting each question correct. The probability that he passes is the probability that he gets 10 correct+probability that he gets 9 correct+probability that he gets 8 correct: P(passes)=P(10 right)+P(9 right)+P(8 right)=[(1/2)^10]+[(1/2)^10]*10+[(1/2)^10]*Combinations(10,2)=[(1/2)^10](1+10+45)=56/1024=7/128.
For Gods sake man, atleast spell your Questions correctly!
100%
4 questions
It is 0.3438
You must first look at all possible outcomes:1, 2, 3, 4, 5, 6Then, reread the question that is being asked:A dice is rolled 6 times, what is the probability of getting atleast a 3.You should underline the important parts of the question and then preceed to answering:When the question states, atleast a 3, it means to say the minimum result of 3 (as in 3, 4, 5 and 6)The math:(Outcomes that satisfy the statement)/(Total outcomes) * 100We multiply by 100 to receive a percentage.4/6 * 100 = 66.67%The answer:The probability of getting atleast a 3 when a dice is rolled 6 times is 66.67%.
The difference between "at least" and "at most" is not restricted to probability. The difference is simply one between the precise meaning of the phrases in every day English language.
1. a) A box of 30 diodes is known to contain five defective ones. If two diodes are selected at random without replacement, what is the probability that atleast one of these diodes is defective
you people are bums answer this question atleast
Ans : Probability that there will be at least one case of twins = 1-(79/80)^20
your question simply makes no sense, or atleast i dont get it.....