Since Luis can only pass or fail his test, and we know that the probability of all possible outcomes is 1, we therefore see that P(Pass) + P(Fail) = 1*where P(X) = the probability of event X*So if P(Pass) = .33;then P(Fail) = 1 - P(Pass)= 1 - .33= .66 (or 66% chance that Luis will fail)
The probability that a parent carrying a dominant trait will pass that trait on is a 50% chance with each pregnancy
The problem as I understand it is the examinee must pass test 1 and test 2 and test 3. Therefore, the probability of passing all 3 exams is .8*.8*.8 = 0.512.
To find the probability of rolling at least one six with one roll of a pair of dice, we need to know two things. First we need to know how many different possible outcomes there are when we roll the dice one time. We know that there are 36 different possible outcomes. They are 1-1, 1-2, 1-3,..., 6-4, 6-5 and 6-6.We also need to know how many different times at least one six can appear. These will include, 1-6, 2-6, 3-6, 4-6, 5-6, 6-6, 6-1, 6-2, 6-3, 6-4, and 6-5, for a total of 11 possible outcomes with at least one six. (Note that the 6-6 possible outcome is included in the list because it fits the requirement of "at least one six" with a given single pass of the dice.The odds of rolling at least one six in one pass with a pair of dice are 11 in 36. To find the probability of this, we need to divide the 11 by the 36, and this gives us a probability of 11/36 or 0.3055555....Remember that probability is a ratio, and it's the ratio of the number of possible times a given outcome can appear to the total number of possible outcomes that could appear. This ratio will range from zero to one, inclusive. A probability of zero means that a given outcome cannot occur. Like rolling a 13 with one pass of a regular pair of dice. That can't happen, so the probability of it happening is zero.We also understand that if the probability of something happening is one, then that outcome must happen. Like the probability of rolling an odd or an even number. We know that an odd or an even number must appear, so the probability of that happening is one.
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 theabove; P(passing)~ 0.1719 ~ 17.2%
Since Luis can only pass or fail his test, and we know that the probability of all possible outcomes is 1, we therefore see that P(Pass) + P(Fail) = 1*where P(X) = the probability of event X*So if P(Pass) = .33;then P(Fail) = 1 - P(Pass)= 1 - .33= .66 (or 66% chance that Luis will fail)
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)!]
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.
Refer back to the first clause. The answer is 50 per cent!
The probability that a parent carrying a dominant trait will pass that trait on is a 50% chance with each pregnancy
what is the probability that exactly 3 students passed the course?
Put in the effort to learn the subjects!
Probability of pass on second attempt is 40% x 80% = 32%
yes
Probability(8 out of 10 pass) = Prob(8 pass and 2 Fail) = (No of ways of selecting 8 out of 10)*0.978*0.032 = 45*0.978*0.032 = 0.0317
Is dehradun university pass out student can apply for goverment sector or public sector?
No. they are not added to the quarterback's statistics.