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Chapter 4: Problem 25

Two coins are to be flipped. The first coin will land on heads with probability \(.6,\) the second with probability \(.7 .\) Assume that the results of the flips are independent, and let \(X\) equal the total number of heads that result. (a) Find \(P\\{X=1\\}\) (b) Determine \(E[X]\)

Short Answer

Expert verified
(a) The probability of getting exactly 1 head when the two coins are flipped is P{X=1} = 0.46. (b) The expected value of the number of heads when the two coins are flipped is E[X] = 1.3.

Step by step solution

01

Probability of Coin 1 landing heads and Coin 2 landing tails

We first calculate the probability of the first coin landing on heads and the second coin landing on tails. P(Coin 1 = Head) = 0.6 P(Coin 2 = Tail) = 1 - P(Coin 2 = Head) = 1 - 0.7 = 0.3 Since the flips are independent, the probability of both events occurring is the product of their individual probabilities. P(Coin 1 = H, Coin 2 = T) = P(Coin 1 = H) * P(Coin 2 = T) = 0.6 * 0.3 = 0.18
02

Probability of Coin 1 landing tails and Coin 2 landing heads

Similarly, calculate the probability of the first coin landing on tails and the second coin landing on heads. P(Coin 1 = Tail) = 1 - P(Coin 1 = Head) = 1 - 0.6 = 0.4 P(Coin 2 = Head) = 0.7 P(Coin 1 = T, Coin 2 = H) = P(Coin 1 = T) * P(Coin 2 = H) = 0.4 * 0.7 = 0.28
03

Probability of Exactly 1 Head

The probability of exactly 1 head (P{X=1}) is the sum of the probabilities calculated above. P{X=1} = P(Coin 1 = H, Coin 2 = T) + P(Coin 1 = T, Coin 2 = H) = 0.18 + 0.28 = 0.46 (b) Expected Value of X (E[X])
04

Probability of Each Possible Outcome

Calculate the probability for each possible number of heads that can result from flipping the coins. P{X=0}: Both coins land tails. P{X=0} = P(Coin 1 = T) * P(Coin 2 = T) = 0.4 * 0.3 = 0.12 P{X=2}: Both coins land heads. P{X=2} = P(Coin 1 = H) * P(Coin 2 = H) = 0.6 * 0.7 = 0.42
05

Expected Value Calculation

Use the probabilities calculated above to find the expected value of X. E[X] = 0 * P{X=0} + 1 * P{X=1} + 2 * P{X=2} E[X] = 0 * 0.12 + 1 * 0.46 + 2 * 0.42 E[X] = 0 + 0.46 + 0.84 = 1.3

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Most popular questions from this chapter

Four buses carrying 148 students from the same school arrive at a football stadium. The buses carry, respectively, \(40,33,25,\) and 50 students. One of the students is randomly selected. Let \(X\) denote the number of students that were on the bus carrying the randomly selected student. One of the 4 bus drivers is also randomly selected. Let \(Y\) denote the number of students on her bus. (a) Which of \(E[X]\) or \(E[Y]\) do you think is larger? Why? (b) Compute \(E[X]\) and \(E[Y]\)

If \(E[X]=1\) and \(\operatorname{Var}(X)=5,\) find (a) \(E\left[(2+X)^{2}\right]\) (b) \(\operatorname{Var}(4+3 X)\)

A gambling book recommends the following "winning strategy" for the game of roulette: Bet \(\$ 1\) on red. If red appears (which has probability \(\frac{18}{38}\) ), then take the \(\$ 1\) profit and quit. If red does not appear and you lose this bet (which has probability \(\frac{20}{38}\) of occurring , make additional \(\$ 1\) bets on red on each of the next two spins of the roulette wheel and then quit. Let \(X\) denote your winnings when you quit. (a) Find \(P\\{X>0\\}\) (b) Are you convinced that the strategy is indeed a "winning" strategy? Explain your answer! (c) Find \(E[X]\)

A ball is drawn from an urn containing 3 white and 3 black balls. After the ball is drawn, it is replaced and another ball is drawn. This process goes on indefinitely. What is the probability that, of the first 4 balls drawn, exactly 2 are white?

Suppose that 10 balls are put into 5 boxes, with each ball independently being put in box \(i\) with probability \(p_{i}, \sum_{i=1}^{5} p_{i}=1\) (a) Find the expected number of boxes that do not have any balls. (b) Find the expected number of boxes that have exactly 1 ball.
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