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katovenus [111]
3 years ago
6

A person invests $1,000 into stock of a company that hopes to go public in one year. The probability that the person will lose a

ll his money after one year (i.e. his stock will be worthless) is 35%. The probability that the person’s stock will still have a value of $1,000 after one year (i.e. no profit and no loss) is 60%. The probability that the person’s stock will increase in value by $10,000 after one year (i.e. will be worth $11,000) is 5%. Find the expected profit after one year.
Mathematics
1 answer:
Lera25 [3.4K]3 years ago
4 0

Answer:

The expected profit after one year is of $150.

Step-by-step explanation:

Probabilities after one year:

35% probability of losing $1,000, that is, 35% probability of a profit of -$1,000.

60% probability of still having a value of $1,000, that is, 60% probability of a profit of $0.

5% probability of a increase in value of $10,000, that is, a profit of $10,000.

Find the expected profit after one year.

Each value is multiplied by its probability. So

E = -0.35*1000 + 0.6*0 + 0.05*10000 = 150

The expected profit after one year is of $150.

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Read 2 more answers
You play two games against the same opponent. The probability you win the first game is 0.4. If you win the first game, the prob
koban [17]

Answer:

a) No, because the probabilities of winning the 2nd game are dependant on the result of the 1st game. The probabilities are different if you win or lose the first game.

b) P=0.42

c) P=0.08

d)

X    |    P(X)

------------------

0    |    0.42

1     |    0.50

2    |    0.08

e) E(x)=0.66

s.d.=0.62

Step-by-step explanation:

a) No, because the probabilities of winning the 2nd game are dependant on the result of the 1st game. The probabilities are different if you win or lose the first game.

b) The probability of losing the first game is

P(G_1=L)=1-P(G_1=W)=1-0.4=0.6

The probability of losing the second game, given that the first game was lost is:

P(G_2=L|G_1=L)=1-P(G_2=W|G_1=L)=1-0.3=0.7

So the probability of losing both games is:

P(G_2=L\&G_1=L)=P(G_1=L)*P(G_2=L|G_1=L)=0.6*0.7=0.42

c) The probability of winning both games is:

P(G_2=W\&G_1=W)=P(G_1=W)*P(G_2=W|G_1=W)=0.4*0.2=0.08

d) The variable X can take values 0, 1 and 2.

X=0 is when both games are lost. This happens with probability P=0.42.

X=2 is when both games are won. This happens with probability P=0.08.

X=1 is when one game is won and the other is lost. This happens with probability P=1-0.42-0.08=0.50.

Then the table of probabilities become:

X    |    P(X)

------------------

0    |    0.42

1     |    0.50

2    |    0.08

e) The expected value is:

E(x)=\sum p_ix_i=0.42*0+0.50*1+0.08*2=0.00+0.50+0.16=0.66

The variance and standard deviation of x are:

V(x)=\sum p_i(x_i-E(x))^2\\\\V(x)=0.42(0-0.66)^2+0.50(1-0.66)^2+0.08(2-0.66)^2\\\\V(x)=0.42*0.4356+0.50*0.1156+0.08*1.7956\\\\V(x)=0.183+0.058+0.144=0.385\\\\\\\sigma=\sqrt{V(x)}=\sqrt{0.385}=0.62

The standard deviation can

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Answer:

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Step-by-step explanation:

(x + 4)(x - 2) = 0 \\ therefore \: x =  - 4 \: (or) \: x =  2

<em>HAVE A NICE DAY</em><em>!</em>

<em>THANKS</em><em> </em><em>FOR GIVING ME THE OPPORTUNITY TO</em><em> </em><em>ANSWER YOUR QUESTION</em><em>. </em>

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