A fair dice is rolled twice. The probability that an odd number will follow an even number is
Here the sample space S = 6
Therefore P(odd number ) = 3/6 = 1/2
and P (even number ) = 3/6 = 1/2
since events are independent,
therefore, P (odd / even )
Three companies X, Y and Z supply computers to a university. The percentage of computers supplied by them and the probability of those being defective are tabulated below
Given that a computer is defective, the probability that it was supplied by Y is
Probability of defective computer supplied by Y =
(Case when Y produces defective)/(All cases of producing defective product)
Case when Y produces defective = (0.3)(0.02) = 0.006
All cases of producing defective product= (0.6x0.01)+(0.3x0.02)
(0.1x0.03)= 0.006+0.006+0.003=0.015
Probability = 0.006/0.015=0.4
An examination consists of two papers, Paper 1 and Paper 2. The probability of failing in Paper 1 is 0.3 and that in Paper 2 is 0.2. Given that a student has failed in Paper 2, the probability of failing in Paper 1 is 0.6. The probability of a student failing in both the papers is
Let A be the event that ‘failed in paper 1’.
B be the event that ‘failed in paper 2’.
Given P(A) = 0.3, P(B) = 0.2.
A fair coin is tossed 10 times. What is the probability that ONLY the first two tosses will yield heads?
Let A be the event that first toss is head
And B be the event that second toss is head.
By the given condition rest all 8 tosses should be tail
∴ The probability of getting head in first two cases
Consider the continuous random variable with probability density function
The standard deviation of the random variables is
T being the random variable of f(t).
The probability that two friends share the same birthmonth is
Let A = the event that the birth month of first friend
And B= that of second friend.
∴ P( A )= 1, as 1st friend can born in any month
and P(B) = 1/12, by the condition.
∴ Probability of two friends share same birthmonth
A box contains 5 block balls and 3 red balls. A total of three balls are picked from the box one after another, without replacing them back. The probability of getting two black balls and one red ball is
Here the possible combination of picking up three balls without replacement is BBR, BRB, RBB.
(B = Black ball, R = Red balls)
∴ Probability of getting two black balls and one red ball is 5/28.
Two dice are thrown. What is the probability that is the sum of the numbers on the two dice is eight?
Here sample space = 6 × 6 = 36
Here, there are five such points whose sum is 8. They are (2,6), (3,5), (4,4), (5,3), (6,2).
Arrivals at a telephone booth are considered to be poison, with an average time of 10 minutes between successive arrivals. The length of a phone call is distributes exponentially with mean 3 minutes. The probability that an arrival does not have to wait before service is
A box contains 5 black and 5 red balls. Two balls are randomly picked one after another from the box, without replacement. The probability for both balls being red is
The probability of drawing two red balls without replacement
A lot has 10% defective items. Ten items are chosen randomly from this lot. The probability that exactly 2 of the chosen items are defective is
Let A be the event that items are defective and B be the event that items are non defective
∴ P( A )= 0.1 and P(B) = 0.9
∴ Probability that exactly two of those items are defective
A box contains 20 defective items and 80 nondefective items. If two items are selected at random without replacement, what will be the probability that both items are defective?
The probability of defective items = 20/100
Therefore the probability of first two defective items without replacement
If three coins are tossed simultaneously, the probability of getting at least one head is
Here the sample space S =2^{3}= 8.
No. of ways to get all tails =1.
∴ probability to get all tails = 1/8
∴ Probability to get at least one head is = 11/8= 7/8
If 20 per cent managers are technocrats, the probability that a random committee of 5 managers consists of exactly 2 technocrats is
The probability of technocrats manager = 20/100 = 1/5
∴ Probability of non technocrats manager = 4/5
Now the require probability
Four arbitrary point (x1,y1), (x2,y2), (x3,y3), (x4,y4), are given in the x, y – plane Using the method of least squares, if, regressing y upon x gives the fitted line y = ax + b; and regressing y upon x given the fitted line y = ax + b; and regressing x upon y gives the fitted line x = cy + d then
y =ax+b − (i) and x = cy + d − (ii)
Chebyshev’s inequality states that the probability of a “Six Sigma” event is less than :
If a bell curve is assumed, the probability of a “six sigma” event is on the order of one ten millionth of a percent.
Three values of x and y are to be fitted in a straight line in the form y = a + bx by the method of least squares. GivenΣx = 6, Σy = 21, Σx^{2} = 14 and Σxy = 46, the values of a and b are respectively.
A box contains 10 screws, 3 of which are defective. Two screws are drawn at random with replacement. The probability that none of the two screws is defective will be
A = First drawn screw nondefective.
B = Second drawn screw nondefective.
P(A) = 7/10
Because 7 out of 10 screws are nondefective and we sample at random,
P(B) = 7/10
The second drawing is the same as the beginning.
P(A Intersection B) = P(A) * P(B) = 0.7 * 0.7 = 0.49
It is Converted to percentage = 0.49 * 100 = 49%
The probability of two screws are defective is 49%.
Which one of the following statements is NOT true?
If the standard deviation of the spot speed of vehicles in a highway is 8.8 kmph and the mean speed of the vehicles is 33 kmph, the coefficient of variation in speed is
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