Answer:
a) 1/2
b) 1/n
c) 1/4
Step-by-step explanation:
a) For each permutation, either 1 precedes 2 or 2 precedes 1. For each permutation in which 1 precedes 2, we can swap 1 and 2 to obtain a permutation in which 2 preceds 1. Thus, half of the total permutations will involve in 1 preceding 2, hence, the probability for a permutation having 1 before 2 is 1/2.
c) If 2 is at the start of the permutation, then it is impossible for 1 to be before 2. If that is not the case, then 1 has a probability of 1/n-1 to be exactly in the position before 2. We can divide in 2 cases using the theorem of total probability,
P( 1 immediately preceds 2) = P (1 immediately precedes 2 | 2 is at position 1) * P(2 is at position 1) + P(1 immediately precedes 2 | 2 is not at position 1) * P(2 is not at position 1) = 0 * 1/n + (1/n-1)*(n-1/n) = 1/n.
d) We can divide the total of permutations in 4 different groups with equal cardinality:
- Those in which n precedes 1 and n-1 precedes 2
- those in which n precedes 1 and 2 precedes n-1
- those in which 1 precedes n and n-1 precedes 2
- those in which 1 precedes n and 2 precedes n-1
All this groups have equal cardinality because we can obtain any element from one group from another by making a permutations between 1 and n and/or 2 and n-1.
This means that the total amount of favourable cases (elements of the first group) are a quarter of the total, hence, the probability of the event is 1/4.
The answer is D, 34
If 4 of the packages contain 7 pens, that means there, 28 pens are total.
Because they contained either 3 or 7 pens, I would assume the remaining 2 packages of 6 total that Steve had bought would have to have 3 pens.
3*2 is equal to 6
28 + 6 = 34
Answer:
ill talk
Step-by-step explanation:
Answer:
30 I am pretty sure
Step-by-step explanation:
Answer:
x=-4
Step-by-step explanation:
-4x -7 - 3x + 4 = 25
-7x=25-4+7
x=-28/7
x=-4
