Can someone help me with this PROBABILITY question? Please.
1 Answer
(i)
Explanation:
The number of ways of choosing
#n * (n - 1) * ... * (n - (k-1)) = (n!)/((n-k)!) = ""^nP_k#
Having chosen
Hence if the order did not matter, then the number of ways of choosing
#""^nC_k = (""^nP_k) / (k!) = (n!)/((n-k)! k!)#
So in the given example we have:
(i)
The number of ways in which
#""^9C_4 = (9!)/((9-4)! 4!) = (9 * 8 * 7 * 6)/(4 * 3 * 2 * 1) = 3024/24 = 126#
(ii)
The number of ways of choosing
#""^4C_2 = (4!)/((4-2)!2!) = (4 * 3) / (2 * 1) = 6#
The number of ways of choosing
#""^3C_1 = (3!)/((3-1)!1!) = (3 * 2) / (2 * 1) = 3#
The number of ways of choosing
#""^2C_1 = (2!)/((2-1)! 1!) = 2 / 1 = 2#
Each of these choices is independent. So the total number of ways of choosing
#""^4C_2 * ""^3C_1 * ""^2C_1 = 6 * 3 * 2 = 36#
(iii)
We know that the unrestricted number of ways of choosing
Of those possibilities, any that miss out one or two composers is unacceptable. Let us attempt to count the unacceptable combinations.
If Beethoven is missed, then there are
#""^5C_4 = (5!)/((5-4)!4!) = 5" "# unacceptable combinations.
If Handel is missed, then there are
#""^6C_4 = (6!)/((6-4)! 4!) = 15" "# unacceptable combinations.
If Sibelius is missed, then there are
#""^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
So the total number of unacceptable combinations is:
#5+15+42-1 = 61#
That leaves the number of acceptable combinations as:
#126-61 = 65#