The probability of getting all 10 questions right is, P(10) =
(1/2)10 =
0.0009765...
9 questions right, P(9) =
10C9 ∙(1/2)10 =
10∙(1/2)10 ~ 0.009765...
8 questions right, P(8) =
10C8 ∙(1/2)10 ~ 45∙(1/2)10 ~ 0.043945...
7 questions right, P(7) =
10C7 ∙(1/2)10 ~ 120∙(1/2)10 ~ 0.1171875...
The probability of passing the test with any of the grade is the sum of all the
above; P(passing)
~ 0.1719 ~ 17.2%
What is the probability of what?Guessing them all correctly?Getting half of the correct?Getting them all wrong?PLEASE be specific with your questions if you want WikiAnswers to help.
The sample space consists of the letters of the word "PROBABILITY" = {P,R,O,B,A,I,L,T,Y}
The probability of a woman being chosen in this situation is 7/11.
The answer depends on what the experiment consists of. Since that is not specified, there can be no sensible answer.
7/11
What is the probability of what?Guessing them all correctly?Getting half of the correct?Getting them all wrong?PLEASE be specific with your questions if you want WikiAnswers to help.
lol
The exam consists of 18 questions and you must answer 15 correctly in order to pass.
If the family consists of a pair of parents and their offspring, then the probability is very close to 0.
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.
If an examination paper has 10 questions and consists of six question in algebra, the other four questions could be geometry, calculus, or trigonometry.
Since there are only two options for the answer, on average the student will answer half of the answers correctly.
The sample space consists of the letters of the word "PROBABILITY" = {P,R,O,B,A,I,L,T,Y}
The probability of a woman being chosen in this situation is 7/11.
The probability histogram consists of eight bars, labeled '1' through '8', each bar having a value (height) of 0.125 .
7/11Note:Choosing people in a committee is not a matter of chance or probability. Personality, skill, seniority etc. enter into the decision.
If there are four possible answers to a question, then a guessed answer would have a probability of 1 in 4. If there are six questions, then the mean number of correct answers would be six times 1 in 4, or 1.5