There are 12 months to choose from There are 7 months with 31 days in them.
The probability of choosing a 31-day month is 7/12.
There are 30 students total. 10 of them have brown hair. 10/30 equals 1/3. Answer: 1/3
What is the third question in the four-way test?
The questions on the four-way test are:
What is the probability of drawing a heart or club from a pack of cards?
Probability of drawing a heart: 1/4
Probability of drawing a club: 1/4
Probability of drawing a heart or a club: 1/4 + 1/4 = 2/4 = 1/2
There are 676,000 ways to make the license plates.
What is the probability that exactly 2 have some kind of defect?
The probability is determined by the binomial distribution. We consider p = probability of defect, q = probability of not defect, n = sample size, and x= number of defects in sample, in this case x=2. We calculate the probability as P(X = x) = n!/[(n-x)! x!] pxqn-x If sample size = 10 and p = 0.1 then: P(x= 2) = 10!/(8!x2!)(0.1)2(0.9)8 = 0.1937 You can find more about the binomial distribution under Wikipedia. It is important also to note the assumptions when using this distribution. It must be a random sample and the probability of defects is known.
1/2 or 50%
Children's genders are not independent and therefore there is no simple answer.
However, if you do assume that their genders are independent then the probability is 1/3.
Is the limit exists for a monotone sequence An?
If a monotone sequence An is convergent, then a limit exists for it. On the other hand, if the sequence is divergent, then a limit does not exist.
The probability is (4/5)3 = 64/125
These are independent one has no bearing on the other
1/4 + 1/8 + 1/2 = 7/8
26 is 1/8th of the balls.
Therefore the total number of balls is 208.
and 1/8th of the balls is 26, so there are 26 blue balls, 104 are yellow and 52 are green.
Because we are only modeling one event, all six outcomes of the die are equally possible. The probability of rolling a four (or, for that matter, any number) is 1/6, or .166666 repeating. Now, since we are modeling 120 rolls, the theoretical number of outcomes of four (or, again, any number) is 1/6 * 120 = 20 outcomes. The second sentence of the problem is unnecessary.
It is 1/593775
What is the probability of rolling a 2 on the first die or the second die?
10 out of 36 or approx 27.78 % (see below for explanation)
If you roll two dice the probability of rolling two on the first OR the second (but not both) is the probability of rolling a two (at least one time) on two throws minus probability of rolling two on both.
Probability of a two coming up at least once is 11 out of 36 (11/36)
You cannot just add 1/6 + 1/6 = 1/3
If that were true then you could throw it six times and get a probability of [1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 = 1] meaning it's guaranteed to happen.
So to get the 11/36, you can create a chart:
11 21 31 41 51 61
12 22 32 42 52 62
13 23 33 43 53 63
14 24 34 44 54 64
15 25 35 45 55 65
16 26 36 46 56 66
And see that it happens 11 out of the possible 36 times, or use a formula.
1 - (probability it won't happen on one try)^(number of tries)
probability it won't happen on one try: 1 - 1/6 = 5/6
So (5/6)^2 = 25/36. Then 1 - 25/36 = 11/36.
Now the chance you get a two on both throws is 1/36, so subtract 11/36 - 1/36 = 10/36. (this can be seen on the chart as 10 positions where there are only one occurrence of two)
The answer depends on whether or not a vehicle being a truck and a vehicle being yellow are independent. Real life suggests that they are not independent events. Consequently, there is insufficient information to provide a useful answer.
If X = time for car to be ready,
Z = (X - m)/s = (60 - 70)/9 = -1.11... Then Prob(Z < -1.111...) = 0.13326
It is 0.0547
If you roll two fair dice what is the probability you will roll a double?
1/6. With two dice there are 36 possible outcomes six of which are doubles 6/36= 1/6.
What is the probability of a coin tossed 3 times landing heads up 3 times?
The is a binomial probability problem, The chance of heads=chance of tails=.5 The general formula for binomial probability is nCrxp^rq^n-r where p is a success and r is a failure. In this case it is 3C3x.5^3 which is 1x.125 so the answer is .125 Another way to think of this problem is the chance of heads is .5 on one toss. If we throw the coin again, the chance of heads is still .5 since this is an independent even and similarly for the third time. So the chance of 3 heads is .5x.5x.5=.125. A more concrete way to see this is List all possible outcomes in the sample space. This time consuming for large n, but for 3 we can do it. HHH, HTH, HHT, THH, HTT, THT, TTH, TTT so there are 8 or 2^3 possible outcomes. Only one of those is 3 heads so 1/8=.125
3/4 x 5/7
21/28 + 20/28
42/56
0.75
How the normal distribution could be transformed to a standard normal distribution?
You may transform a normal distribution curve, with, f(x), distributed normally, with mean mu, and standard deviation s, into a standard normal distribution f(z), with mu=0 and s=1, using this transform: z = (x- mu)/s