A die is rolled three times. The probability that exactly one odd number turns up among the three outcomes is
Probability of getting an odd number in rolling of a die = 3/6 = 1/2.
Now using binomial distribution
P(Exactly one odd number among three outcomes)
Seven (distinct) car accidents occurred in a week. What is the probability that they all occurred on the same day?
Sample space = 7^{7}
All accidents on the same day = 7 ways (all on Monday, all on Tuesday...)
So, required probability
Let X and Y be two exponentially distributed and independent random variables with mean α and β, respectively. If Z= min (X, Y), then the mean of Z is given by
X and Y are two independent exponentially distributed random variables. Let λ_{1} and λ_{2} parameters of X and Y respectively.
Given, Z = min (X, Y)
Since mean of exponential distribution = 1/Parameter
So,
∴ Z is random variable with parameter
Mean of Z =
A bag contains 10 blue marbles, 20 green marbles and 30 red marbles. A marble is drawn from the bag, its colour recorded and it is put back in the bag. This process is repeated 3 times. The probability that no two of the marbles drawn have the same colour is
The given condition corresponds to sampling with replacement and with order.
No 2 marbles have the same color i.e. Drawn 3 different marble.
So total number of ways for picking 3 different marbles = 3! = 6.
Probability of getting blue, green, red in order
[Since 6 ways to get the marbles]
= 1/6
In a certain town, the probability that it will rain in the afternoon is known to be 0.6. Moreover, meteorological data indicates that if the temperature at noon is less than or equal to 25^{0}C, the probability that it will rain in the afternoon is 0.4. The temperature at noon is equally likely to be above 25^{0}C, or at/below 25^{0}C.
What is the probability that it will rain in the afternoon on a day when the temperature at noon is above 25^{0}C?
Let RA : Rain in the afternoon
T > 25 : Temperature more than 25^{0}C
Let the desired probability = P(RAIT ≤ 25) = x
The tree diagram forthis problem is given below:
Given P(RA) = 0.6
by rule of total probability P(RA)
⇒ x = 0.8
Suppose a fair sixsided die is rolled once, if the value on the die is 1, 2 or 3 then die is rolled a second time. What is the probability that the sum total of values that turn up is at least 6?
If first throw is 1, 2 or 3 then sample space is only 18 possible ordered pairs. Out of this only (1, 5 ), (1, 6 ), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5) and (3,6) i e. 9 out of 18 ordered pairs gives a Sum ≥ 6.
If first throw is 4, 5 or 6 then second throw is not made and therefore the only way Sum ≥ 6 is if the throw was 6. Which is one out of 3 possible. So the tree diagram becomes as follows:
From above diagram
Suppose p is the number of cars per minute passing through a certain road junction between 5 PM and p has Poisson distribution with mean 3. What is the probability of observing fewer than 3 cars during any given minute in this interval?
Probability of observing fewer than 3 cars
A fair coin is tossed till a head appears for the first time. The probability that the number of required tosses is odd, is
P(number of tosses is odd) = P(number of tosses is 1, 3, 5, 7 ...)
P(number of toss is 1) = P(Head in first toss = 1/2
P(number of toss is 3) = P(tail in first toss, tail in second toss and head in third toss)
P(number of toss is 5) = P(T, T, T, T, H)
So P(number of tosses is odd)
Sum of infinite geometric series with
A continuous random variable Xhas a probability density function f(x) = e^{x}, 0 < x <∝. Then P{X> 1} is
A fair coin is tossed n times. The probability that the difference between the number of heads and tails is ( n  3) is
when n = probability of occurrence of head
y = probability of occurrence of tail
Let number of head is P
Number of tail is q
Here required probability is zero.
Two players, A and B, alternately keep rolling a fair dice. The person to get a six first wins the game. Given that player A starts the game, the probability that A wins the game is
P(A wins) = p(6 in first throw by A) + p(A not 6, B not 6, A 6) + ...
Equation e^{x}  1 = 0 is required to be solved using Newton's method with an initial guess x_{0} = 1. Then, after one step of Newton’s method, estimate x_{1} of the solution will be given by
Let x^{2}  117 = 0. The iterative steps for the solution using NewtonRaphson’s method is given by
Solution of the variables x_{1} and x_{2} for the following equations is to be obtained by employing the NewtonRaphson iterative method
Assuming the initial values x_{1} = 0.0 and x_{2} = 1.0, the Jacobian matrix is
When the NewtonRaphson method is applied to solve the equation f(x) = x^{3} + 2x  1 = 0, the solution at the end of the first iteration with the initial guess value as x_{0} = 1.2 is
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