Suppose we want to form shopping lists out of the following three sets of items,

Question

Suppose we want to form shopping lists out of the following three sets of items, so that each list contains precisely one item from each set A_1, A_2, A_3. A_1 = {'water', 'beer', 'wine'}, A_2 = {'chocolate', 'ice cream'}, A_3 = {'apple', 'pear', 'peach', banana'} a) What are the distinct numbers of possible shopping lists we can obtain in this way, assuming that the way we order the items does not matter? b) How does the number of possible shopping lists change if the order the items are written down on the list does matter? Suppose we observe a random sequence of letters consisting of four As, five Bs, eight Cs and three Ds. Under the assumption that all such sequences are equally probable, compute the probability that the random sequence consists of four blocks, with each block containing precisely one of the four letters. Examples: AAAABBBBBCCCCCCCCDDD, DDDBBBBBAAAACCCCCCCC, CCCCCCCCDDDAAAABBBBB, ... In drawing nine cards with replacement from an ordinary deck of 52 cards, what is the probability of drawing three aces of spades? By contrast, What is the probability of obtaining the three aces in precisely the first three draws? A town has five parks. On a Saturday, five classmates, who are unaware of each other's decision, choose a park at random and go there at the same time. What is the probability that at least two of them go to the same park?

Explanation / Answer

Answer to question asked:

0.9917

EXPLANATION:

P(at lest 3 of them go to the same park) = 1 - P (each of the 5 classmates go to diffent parks)         (1)

First classmate can go any of the 5 parks.

So, P(First classmate) = 1/5

Second classmate can go to 4 praks, other than the 1 park First classmate has gone.

So, P(Second classmate) = 1/4

Third classmate can go to 3 parks, other than the 2 parks First & Second classmates have gone.

So, P(Third classmate) = 1/3

Fourth classmate can go to 2 parks, other than the 3 parks First, Second & Third classmates have gone.

So, P(Fourth classmate = 1/2

Fifthclassmate can to the remaining park.

So, P(Fifth classmate) = 1

Thus, P(each of the 5 classmates go to different parks) = 5X4X3X2X1 = 1/120 = 0.0083

So, P(at least two them go to the same park) = 1 - 0.0083 = 0.9917

Problem 1:

(a)

1 item from A1 can be chosen in = 3 ways (since there are 3 items in A1)

1 item from A2 can be chosen in = 2 ways (since there are 2 items in A2)

1 item from A3 can be chosen in = 4 ways (since there are 4 items in A3)

So, total number of ways = 3 X 2 X 4 = 24 ways

(b)

Since A1, A2 and A3 can be arranged in = 3! = 6 ways,

Total number of ways = 6 X 24 = 144

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