A box contains 11 red chips and 4 blue chips. We perform the following two-step experiment: (1) First, a chip is selected at random from the box and is then removed from the box. (After this first step, there are 14 chips left in the box. ) (2) Then, a chip is selected at random from the box (that is, from the remaining 14 chips) . Let B 1 be the event that the chip removed from the box at the first step of the experi- ment is red. Let B 2 be the event that the chip removed from the box at the first step of the experiment is blue. Let A be the event that the chip selected from the box at the second step of the experiment is red. Find P(B1), P(B2), P(A), P(B1|A), and P(B2|A).

P(B1) = (11/15)

P(B2) = (4/15)

P(A) = (11/15)

P(B1|A) = (5/7)

P(B2|A) = (2/7)

Step-by-step explanation:

There are 11 red chips and 4 blue chips in a box. Two chips are selected one after the other at random and without replacement from the box.

B1 is the event that the chip removed from the box at the first step of the experiment is red.

B2 is the event that the chip removed from the box at the first step of the experiment is blue. A is the event that the chip selected from the box at the second step of the experiment is red.

Note that the probability of an event is the number of elements in that event divided by the Total number of elements in the sample space.

P(E) = n(E) ÷ n(S)

P(B1) = probability that the first chip selected is a red chip = (11/15)

P(B2) = probability that the first chip selected is a blue chip = (4/15)

P(A) = probability that the second chip selected is a red chip

P(A) = P(B1 n A) + P(B2 n A) (Since events B1 and B2 are mutually exclusive)

P(B1 n A) = (11/15) × (10/14) = (11/21)

P(B2 n A) = (4/15) × (11/14) = (22/105)

P(A) = (11/21) + (22/105) = (77/105) = (11/15)

P(B1|A) = probability that the first chip selected is a red chip given that the second chip selected is a red chip

The conditional probability, P(X|Y) is given mathematically as

P(X|Y) = P(X n Y) ÷ P(Y)

So, P(B1|A) = P(B1 n A) ÷ P(A)

P(B1 n A) = (11/15) × (10/14) = (11/21)

P(A) = (11/15)

P(B1|A) = (11/21) ÷ (11/15) = (15/21) = (5/7)

P(B2|A) = probability that the first chip selected is a blue chip given that the second chip selected is a red chip

P(B2|A) = P(B2 n A) ÷ P(A)

P(B2 n A) = (4/15) × (11/14) = (22/105)

P(A) = (11/15)

P(B2|A) = (22/105) ÷ (11/15) = (2/7)

Hope this Helps!!!

introduction, methodology, results, discussion

step-by-step explanation:

an observational study is one which consists of actual observation to record data and then analyse the data.

there are a number of factors that should be analyzed before starting an observational study.

the main four factors are:

i) introduction:   this should be analysed to check whether the purpose of the study is correctly established and the data recorded are appropriate for the study

methodology:   the method of collecting the data, such as whether randomly selected, and whether sample size matches with the population to be studied, whether questions asked are easy to answer by common men, etc.

results: then results such as summary of data with comments should be studied thoroughly

discussion:   the last is the discussion, this can be either by management people or persons who are all interested , etc.

total amount of money that he should pay if he has not used his duty free exemption is $98

step-by-step explanation:

given a united states citizen returning to the states declares the following items at the customs office: 3 shirts at $8.50 each 2 dresses at $27.50 each 1 pair of gold cuff links at $17.50 per pair. we have to find how much he pay

cost of 3 shirts if one cost at $8.50 is

cost of 2 dresses if one cost at $27.50 is  

cost of 1 pair of gold cuff at $17.50

so, total amount of money that he should pay if he has not used his duty free exemption is   $25.5+$55+$17.50=$98