26. The law of permutations, combinations, and filling slots are the counting rules used to
(a) determine the total number of possible outcomes
(b) determine all possible combinations of “n” objects from a total of N objects
(c) determine the odds of winning a lottery
(d) calculate probabilities using different probability approaches
27. Two red balls and one black ball from a total of 6 red and 4 black balls can be drawn in
the following way:
(a)
6 4
2 0 C XC
(b )
6 4
2 1 C XC
(c)
(d)
6 4
4 1 C XC
6 4
26. The law of permutations, combinations, and filling slots are the counting rules used to:
(a) determine the total number of possible outcomes
(b) determine all possible combinations of "n" objects from a total of N objects
(c) determine the odds of winning a lottery
(d) calculate probabilities using different probability approaches
To apply the counting rules, we need to understand the concepts of permutations, combinations, and filling slots.
Permutations: Permutations refer to the arrangement of objects in a particular order. The formula for calculating permutations is P(n, r) = n! / (n - r)!, where n is the total number of objects, and r is the number of objects being arranged in a specific order.
Combinations: Combinations refer to the selection of objects without considering the order in which they are arranged. The formula for calculating combinations is C(n, r) = n! / (r! * (n - r)!), where n is the total number of objects, and r is the number of objects being selected.
Filling Slots: Filling slots involves allocating objects to different positions or slots. The formula for filling slots is XC(n1, r1) * XC(n2, r2) * XC(n3, r3)..., where XC represents the ways to fill slots, n1, n2, n3... represent the available objects for each slot, and r1, r2, r3... represent the number of slots to be filled.
Now let's look at the options for question 27:
(a) 6 4
2 0 C XC
This option represents selecting 2 red balls and 0 black balls from a total of 6 red balls and 4 black balls. Using the combination formula, C(n, r), we can calculate it as C(6, 2) * C(4, 0) = 15 * 1 = 15. So, there are 15 ways to select 2 red balls without any black balls.
(b) 6 4
2 1 C XC
This option represents selecting 2 red balls and 1 black ball from a total of 6 red balls and 4 black balls. Using the combination formula, C(n, r), we can calculate it as C(6, 2) * C(4, 1) = 15 * 4 = 60. So, there are 60 ways to select 2 red balls and 1 black ball.
(c) 6 4
4 1 C XC
This option represents selecting 4 red balls and 1 black ball from a total of 6 red balls and 4 black balls. Using the combination formula, C(n, r), we can calculate it as C(6, 4) * C(4, 1) = 15 * 4 = 60. So, there are 60 ways to select 4 red balls and 1 black ball.
(d) 6 4
It seems there is incomplete information provided for this option. Please provide the rest of the question for a proper explanation.
If you have any further questions, feel free to ask!