Probability
Three coins are tossed up in the air, one at a time. What is the probability that two of them will land heads up and one will land tails up?
is it 3/8?: {HHH HHT HTT HTH TTT TTH THT THH}
Using a six-sided die, Carlin has rolled a six on each of 4 successive tosses. What is the probability of Carlin rolling a six on the next toss?
Is it 1/6?
A fair coin is tossed twice. Let X be the number of heads that are observed. Construct the probability distribution of X. Find the probability that at least one head is observed.
x = 0, 1, 2
Px=0.25, 0.5, 0.25
P(X≥1)=P(1)+P(2)=0.50+0.25=0.75
Find the mean of the discrete random variable X whose probability distribution is
x= -2, 1, 2, 3.5
P(x)= 0.21, 0.34.0.24, 0.21
μ=Σx P(x)=(−2)⋅0.21+(1)⋅0.34+(2)⋅0.24+(3.5)⋅0.21=1.135
A regular deck of cards has 52 cards. Assuming that you do not replace the card you had drawn before the next draw, what is the probability of drawing three aces in a row?
Is it 1/5,525?
P=1/13*1/17*1/25=1/5,525
In a sample of 40 vehicles, 18 are red, 6 are trucks, and 2 are both. Suppose that a randomly selected vehicle is red. What is the probability it is a truck?
1/9
P(truck|red)=P(truck andred)/P(red)=(2/40)/(18/40)=2/18=1/9
A pair of fair dice is rolled. Let X denote the sum of the number of dots on the top faces.
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
The possible values for X are the numbers 2 through 12. X = 2 is the event {11}, so P(2)=1∕36. X = 3 is the event {12,21}, so P(3)=2∕36. So on.
P(X≥9)=P(9)+P(10)+P(11)+P(12)=4/36+3/36+2/36+1/36=10/36=0.27
P(X is even)=P(2)+P(4)+P(6)+P(8)+P(10)+P(12)=1/36+3/36+5/36+5/36+3/36+1/36=18/36=0.5
In a hamster breeder's experience the number X of live pups in a litter of a female not over twelve months in age who has not borne a litter in the past six weeks has the probability distribution
x= 3,4,5,6,7,8,9
P(x)=0.04, 0.10, 0.26, 0.31, 0.22, 0.05, 0.02
0.79
0.60
μ = 5.8, σ = 1.2570
Find the probability of getting two heads when five coins are tossed.
Number of ways of getting two heads = 5C2 = 10. Total Number of ways = 25 = 32
P (two heads) = 10/32 = 5/16
Two dies are thrown simultaneously, and the sum of the numbers obtained is found to be 7. What is the probability that the number 3 has appeared at least once?
Event A indicates the combination in which 3 has appeared at least once.
Event B indicates the combination of the numbers which sum up to 7.
A = {(3, 1), (3, 2), (3, 3)(3, 4)(3, 5)(3, 6)(1, 3)(2, 3)(4, 3)(5, 3)(6, 3)}
B = {(1, 6)(2, 5)(3, 4)(4, 3)(5, 2)(6, 1)}
P(A) = 11/36
P(B) = 6/36
A ∩ B = 2
P(A ∩ B) = 2/36
Applying the conditional probability formula we get,
P(A|B) = P(A∩B)/P(B) = (2/36)/(6/36) = 1/3
A discrete random variable X has the following probability distribution:
x= -1, 0, 1, 4
P(x)= 0.2, 0.5, a, 0.1
Compute the following:
Two fair dice are rolled at once. Let X denote the difference in the number of dots that appear on the top faces of the two dice. Thus for example if a one and a five are rolled, X = 4, and if two sixes are rolled, X = 0.
1. x= 0,1,2,3,4,5
px= 6/36, 10/36, 8/36, 6/36, 4/36, 2/36
2. 1.9444
3. 1.4326
Three dice are rolled together. What is the probability as getting at least one '4'?
= 1 – (Probability of getting no number 4) = 1 – (5/6) x (5/6) x (5/6) = 91/216
A board game comes with a special deck of cards, some of which are black, and some of which are gold. If a card is randomly selected, the probability it is gold is 0.20, while the probability it gives a second turn is 0.16. Finally, the probability that it is gold and gives a second turn is 0.08. Suppose that a card is randomly selected, and it allows a player a second turn. What is the probability it was a gold card?
This time, we are given the following probabilities:
We are trying to calculate: P(gold|second turn)
We can apply the formula to find this probability:
P(gold|second turn)=
P(gold and second turn)/P(second turn)=0.08/0.16=0.5
A service organization in a large town organizes a raffle each month. One thousand raffle tickets are sold for $1 each. Each has an equal chance of winning. First prize is $300, second prize is $200, and third prize is $100. Let X denote the net gain from the purchase of one ticket.
If a ticket is selected as the first prize winner, the net gain to the purchaser is the $300 prize less the $1 that was paid for the ticket, hence X = 300 − 1 = 299. There is one such ticket, so P(299) = 0.001. Applying the same “income minus outgo” principle to the second and third prize winners and to the 997 losing tickets yields the probability distribution:
x= 299, 199, 99, −1
P(x)0.001, 0.001, 0.001, 0.997
Let W denote the event that a ticket is selected to win one of the prizes. Using the table
P(W)=P(299)+P(199)+P(99)=0.001+0.001+0.001=0.003
A life insurance company will sell a $200,000 one-year term life insurance policy to an individual in a particular risk group for a premium of $195. Find the expected value to the company of a single policy if a person in this risk group has a 99.97% chance of surviving one year.
et X denote the net gain to the company from the sale of one such policy. There are two possibilities: the insured person lives the whole year or the insured person dies before the year is up. Applying the “income minus outgo” principle, in the former case the value of X is 195 − 0; in the latter case it is 195−200,000=−199,805.195−200,000=−199,805. Since the probability in the first case is 0.9997 and in the second case is 1−0.9997=0.00031−0.9997=0.0003, the probability distribution for X is:
x= 195, −199,805
P(x)=0.9997, 0.0003
E(X)=Σx P(x)=195⋅0.9997+(−199,805)⋅0.0003=135