This can be solved by finding the number of permutations of the eight performers who will all perform before the one who insists on being the last.
Identify the center and radius of the circle with equation. ( please help me )
1 answer:
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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
Start with a deck of 52 cards. Draw 1 card. Probability of drawing a diamond first:
Now draw another single card from the deck, which now contains 51 cards. Probability of drawing a spade:
Draw 1 last card. The deck has 50 cards left, and 12 of these are diamonds. Probability of drawing a diamond:
So the probability of drawing a diamond, then a spade, then another diamond is
Now draw another single card from the deck, which now contains 51 cards. Probability of drawing a spade:
Draw 1 last card. The deck has 50 cards left, and 12 of these are diamonds. Probability of drawing a diamond:
So the probability of drawing a diamond, then a spade, then another diamond is