Canadians who visit the United States often buy liquor and cigarettes, which are much cheaper in the United States. However, there are limitations. Canadians visiting in the United States for more than 2 days are allowed to bring into Canada one bottle of liquor and one carton of cigarettes. A Canada Customs agent has produced the following joint probability distribution of the number of bottles of liquor and the number of cartons of cigarettes imported by Canadians who have visited the United States for 2 or more days.

Bottle liquor
Carton cigarettes 0 1
0 0.62 0.16
1 0.07 0.15

Find the marginal probability distribution of the number of bottles imported.

Probability of bringing a bottle of liquor into the country that is, the probability of bringing 1 bottle liquor into the country = P(B) = 0.31

The probability of not bringing a bottle of liquor into the country, that is, the probability of bringing 0 bottle liquor into the country = P(B') = 0.69

Probability distribution of bottle liquor

Let X represent the random variable of the number of bottle liquor brought into the country by a person

X | P(X)

0 | 0.69

1 | 0.31

Step-by-step explanation:

The joint probability distribution for the number of bottles of liquor and the number of cartons of cigarettes imported by Canadians who have visited the United States for 2 or more days is given in the question as

V | B

C | 0 | 1

0 | 0.62 | 0.16

1 | 0.07 | 0.15

Note that B = bottle liquor

C = Carton cigarettes

V is each variable

Let the probability of bringing a bottle of liquor into the country be P(B), that is, the probability of bringing 1 bottle liquor into the country.

The probability of not bringing a bottle of liquor into the country is P(B'), that is, the probability of bringing 0 bottle liquor into the country.

Let the probability of bringing a carton of cigarettes into the country be P(C), that is, the probability of bringing 1 carton cigarettes into the country.

The probability of not bringing a carton of cigarettes into the country is P(C'), that is, the probability of bringing 0 carton cigarettes into the country.

From the joint probability table, we can tell that

P(B n C) = 0.15

P(B n C') = 0.16

P(B' n C) = 0.07

P(B' n C') = 0.62

Find the marginal probability distribution of the number of bottles imported.

Probability of bringing a bottle of liquor into the country that is, the probability of bringing 1 bottle liquor into the country = P(B)

P(B) = P(B n C) + P(B n C') = 0.15 + 0.16 = 0.31

The probability of not bringing a bottle of liquor into the country, that is, the probability of bringing 0 bottle liquor into the country = P(B')

P(B') = P(B' n C) + P(B' n C') = 0.07 + 0.62 = 0.69

Probability distribution of bottle liquor

Let X represent the random variable of the number of bottle liquor brought into the country by a person

X | P(X)

0 | 0.69

1 | 0.31

Hope this Helps!!!

a framing bias

step-by-step explanation:

rodrigo is subject to a framing bias i believe.