P(X|Y) = P(Y intersection X) / P(Y); where
P(X|Y) is probability of event X provided event Y had already occurred
P(Y) is probability of event Y happening
P(Y intersection X) is probability of events Y & X occurring together
Q3.a
P(Y): Prob of at least one insurance schemes (A or B) has been sold
= 1 - Prob of none of the schemes sold
= 1 - (1-Prob of A being Sold)*(1-Prob of B being Sold) // As schemes A & B are independent
= 1 - (1-0.6)(1-0.4)
= 1 - (0.4)*(0.6)
= 1 - 0.24
= 0.76
P(X intersection Y): Prob of at least one insurance schemes (A or B) has been sold AND also scheme 'A' being sold
= Prob(A sold) and Prob (B sold) + Prob(A sold) and Prob (B not sold)
= 0.6 * 0.4 + 0.6 * (1-0.4)
= 0.24 + 0.36
= 0.6
P(X|Y): Prob scheme A has been sold given that at least one insurance scheme has been sold
= P(X intersection Y) / P(Y)
= 0.6 / 0.76
= 15/19
Since having a child to a child is an independent event (assuming no outside intervention), the probability is still about 50 / 50 boy or girl.
Assuming that having boys and girls are equally likely, then the probability is 1/8. * * * * * You also need to assume that the children's genders are independent. They are NOT and depend on the parents' ages and genes.
For a four digit pin number: You receive the first PIN number, let's say WXYZ. The probability that the next pin number you receive would match (assuming they are randomly provided), is: For each digit, they are 10 possibilities [0 1 2 3 4 5 6 7 8 8]. The probability that one specific number is chosen is thus of 1/10. For the are four digits, hence four independent selection of one digit, each with a probability of 1/10. The probability of an event, combination of independent events, is the product of the the probability of the independent event. Thus, the probability that the next pin number you receive would match (assuming they are randomly provided), is: 1/10*1/10*1/10*1/10 or 1/10^4 or 0.0001 or 1 out 10000
Assuming the probability of a boy is 0.5, the probability of a boy and boy and boy is 0.5 * 0.5 * 0.5 = 0.125.
Assuming that "piossion" refers to Poisson, they are simply different probability distributions that are applicable in different situations.
Since having a child to a child is an independent event (assuming no outside intervention), the probability is still about 50 / 50 boy or girl.
Assuming that having boys and girls are equally likely, then the probability is 1/8. * * * * * You also need to assume that the children's genders are independent. They are NOT and depend on the parents' ages and genes.
Assuming the probability of having a boy is 1/2, and that the probabilities are independent: Probability of 1 girl and 12 boys = (1/2)13 * 13 = 0.001587..., which is around 1/630
50%....maybe you're not cut out for college....
Assuming that probility is your failed attempt to spell probability, the activity is repeated trials whose outcomes are independent, identically distributed (iid) variables.
For a four digit pin number: You receive the first PIN number, let's say WXYZ. The probability that the next pin number you receive would match (assuming they are randomly provided), is: For each digit, they are 10 possibilities [0 1 2 3 4 5 6 7 8 8]. The probability that one specific number is chosen is thus of 1/10. For the are four digits, hence four independent selection of one digit, each with a probability of 1/10. The probability of an event, combination of independent events, is the product of the the probability of the independent event. Thus, the probability that the next pin number you receive would match (assuming they are randomly provided), is: 1/10*1/10*1/10*1/10 or 1/10^4 or 0.0001 or 1 out 10000
Assuming the probability of a boy is 0.5, the probability of a boy and boy and boy is 0.5 * 0.5 * 0.5 = 0.125.
Assuming that "piossion" refers to Poisson, they are simply different probability distributions that are applicable in different situations.
assuming that it is a regular die with numbers 1-6 on them the probability is 1
Assuming that the colors are balanced, the probability is 1 in 5.
assuming a single six sided die the probability of rolling a three is 1/6
Assuming that you mean David Beckham, the probability is very close to 0.