ANSWER:
E[X] ≈ m ln m
STEP-BY-STEP EXPLANATION:
Hint: Let X be the number needed. It is useful to represent X by
m
X = ∑ Xi
i=1
where each Xi is a geometric random variable
Solution: Assume that there is a sufficiently large number of coupons such that removing a finite number of them does not change the probability that a coupon of a given type is draw. Let X be the number of coupons picked
m
X = ∑ Xi
i=1
where Xi is the number of coupons picked between drawing the (i − 1)th coupon type and drawing i th coupon type. It should be clear that X1 = 1. Also, for each i:
Xi ∼ geometric 
P r{Xi = n} =
Such a random variable has expectation:
E [Xi
] =
= 
Next we use the fact that the expectation of a sum is the sum of the expectation, thus:
m m m m
E[X] = E ∑ Xi = ∑ E Xi = ∑ = m ∑ 
= mHm
i=1 i=1 i=1 i=1
In the case of large m this takes on the limit:
E[X] ≈ m ln m
The odds of two students getting the same number, regardless of which number the first student gets, is 1:25. The odds of them getting the same number--such as both of them getting 13--would be 25*25=1:625. These are the same odds of three students getting the same numbers in a row without considering which number they get in a row. Then, if you add another student, just multiply the previous answer by 25.
Answer:
2(x + 1) = 10
2x + 2 = 10
2x +2 -2 = 10 - 2
2x = 10(Step 3)
2x/2 = 8/2
x = 4
Step-by-step explanation:
Given the equation:
2(x + 1) = 10
Open the bracket by multiplying 2 by (x + 1) - Distributive property
2x + 2 = 10;
Subtract 2 from both sides.
2x + 2 - 2 = 10 - 2( Subtraction property of equality)
2x = 8(Step 3 justified).
Divide both sides by 2
2x/2 = 8/2 - Division property of equality.
Therefore,
x = 4. (Solved)
C to your question for this
12.6$Answer:
Step-by-step explanation:
1$ - 20$ = 19$ / 3 = 6.3$
6.3$ x 2 = 12.6$
