Consider a game in which a player rolls a number cube to determine the number of points earned. If a player rolls a number great

er than or equal to 4, the number of the roll is added to the total points. Any other roll is deducted from the player’s total. What is the expected value of the points earned on a single roll in this game?

Mathematics

1 answer:

Vanyuwa [196]3 years ago

3 0

$E_{roll}= \sum_{i=1}^{n} (x_i)(p_i)$
$E_{roll}= ((-1)(\frac{1}{6}))+((-2)(\frac{1}{6}))+((-3)(\frac{1}{6}))+((4)(\frac{1}{6}))+((5)(\frac{1}{6}))+((6)(\frac{1}{6}))$
$E_{roll}= (\frac{-1}{6})+(\frac{-1}{3})+(\frac{-1}{2})+(\frac{2}{3})+(\frac{5}{6})+(1)((6)(\frac{1}{6}))$
$E_{roll}= \frac{3}{2}$

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You have a biased coin for which p(h)=pp(h)=p. you toss the coin 2020 times. what is the probability that you observe 88 heads a

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The question describes a binomial probability with p(h) = p, then p(t) = 1 - p and number of trials (n) = 20

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Part A:

The probability of observing 8 heads and 12 tails is given by:

$P(8)=\, ^{20}C_8p^8(1-p)^{20-8}=\, ^{20}C_8p^8(1-p)^{12}$

Part B:

<span>You observe more than 8 heads and more than 8 tails, when you observe 9 heads and 11 tails, 10 heads and 10 tails, and 11 heads and 9 tails.

Therefore, the probability of </span><span>observing more than 8 heads and more than 8 tails</span> is given by:

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