Can someone help me with this PROBABILITY question? Please.

Answer 1

Quinn Allen

(i) #126#; (ii) #36#; (iii) #65#

The number of ways of choosing #k# items from #n# distinct items if the order of the choice matters is:

#n * (n - 1) * ... * (n - (k-1)) = (n!)/((n-k)!) = ""^nP_k#

Having chosen #k# items from #n#, then the number of ways in which those #k# items can be ordered is #k!#.

Hence if the order did not matter, then the number of ways of choosing #k# items from #n# distinct items is:

#""^nC_k = (""^nP_k) / (k!) = (n!)/((n-k)! k!)#

Thus, we have in the example provided:

(i)

The number of ways in which #4# pieces can be chosen from #9# is:

#""^9C_4 = (9!)/((9-4)! 4!) = (9 * 8 * 7 * 6)/(4 * 3 * 2 * 1) = 3024/24 = 126#

(ii)

The number of ways of choosing #2# pieces by Beethoven is:

#""^4C_2 = (4!)/((4-2)!2!) = (4 * 3) / (2 * 1) = 6#

The number of ways of choosing #1# piece by Handel is:

#""^3C_1 = (3!)/((3-1)!1!) = (3 * 2) / (2 * 1) = 3#

The number of ways of choosing #1# piece by Sibelius is:

#""^2C_1 = (2!)/((2-1)! 1!) = 2 / 1 = 2#

Each of these choices is independent. So the total number of ways of choosing #4# pieces as required is their product:

#""^4C_2 * ""^3C_1 * ""^2C_1 = 6 * 3 * 2 = 36#

(iii)

We know that the unrestricted number of ways of choosing #4# pieces from #9# is #""^9C_4 = 126#

Let's try to count the combinations that are unacceptable, starting with the ones that omit one or two composers.

If Beethoven is missed, then there are #5# other pieces, giving us:

#""^5C_4 = (5!)/((5-4)!4!) = 5" "# unacceptable combinations.

If Handel is missed, then there are #6# other pieces, giving us:

#""^6C_4 = (6!)/((6-4)! 4!) = 15" "# unacceptable combinations.

If Sibelius is missed, then there are #7# other pieces, giving us:

#""^7C_4 = (7!)/((7-4)! 4!) = (7 * 6 * 5) / (3 * 2 * 1) = 42" "# unacceptable combinations.

Of these unacceptable combinations, there is one that misses two composers - namely Handel and Sibelius, leaving us with the #4# pieces by Beethoven. That possibility is included twice.

Thus, the total quantity of inappropriate pairings is:

#5+15+42-1 = 61#

Thus, the total number of feasible pairings is as follows:

#126-61 = 65#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.