SOLUTION: An investor invests in stocks without being informed of all the intricacies of stock market. The probability that his 1st investment in one market sector is a good stock is 1/3 and
Algebra ->
Probability-and-statistics
-> SOLUTION: An investor invests in stocks without being informed of all the intricacies of stock market. The probability that his 1st investment in one market sector is a good stock is 1/3 and
Log On
Question 1036314: An investor invests in stocks without being informed of all the intricacies of stock market. The probability that his 1st investment in one market sector is a good stock is 1/3 and that his second investment in an unrelated market sector is a good stock is 1/4.
Find THE probability that:
A)At least one investment would be a good stock
B)Both the investments would be good stocks
C) Only one investment would be a good stock
D)None of the investments would be a good stock.
Thanks in advance! Found 2 solutions by stanbon, jim_thompson5910:Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! n investor invests in stocks without being informed of all the intricacies of stock market. The probability that his 1st investment in one market sector is a good stock is 1/3 and that his second investment in an unrelated market sector is a good stock is 1/4.
Find THE probability that:
A)At least one investment would be a good stock
P(at least one good) = 1 - P(both bad) = 1-(2/3)(3/4) = 1/2
=========================================
B)Both the investments would be good stocks
P(both good) = 1/3 *1/4 = 1/12
-------------------------------------------
C) Only one investment would be a good stock
P(one good) = 1 - [(1/12)+(1/2)] = 1-7/12 = 5/12
------------------------------
D)None of the investments would be a good stock.
P(none good) = (2/3)(3/4) = 1/2
--------------
Cheers,
Stan H.
-----------
An investor invests in stocks without being informed of all the intricacies of stock market. The probability that his 1st investment in one market sector is a good stock is 1/3 and that his second investment in an unrelated market sector is a good stock is 1/4.
Find THE probability that:
A)At least one investment would be a good stock
B)Both the investments would be good stocks
C) Only one investment would be a good stock
D)None of the investments would be a good stock.
==========================================================================
Steps:
Let
X = event that his 1st investment is a good stock
X' = event that his 1st investment is NOT a good stock
Y = event that his 2nd investment is a good stock
Y' = event that his 2nd investment is NOT a good stock
---------------------------------------------
(A)
The two events are independent, so we can use the rule that P(X and Y) = P(X)*P(Y)
The probability that BOTH investments are bad investments is the same as asking for the probability of event X' happening and event Y' happening as well.
P(X' and Y') = P(X')*P(Y')
P(X' and Y') = (2/3)*(3/4)
P(X' and Y') = (2*3)/(3*4)
P(X' and Y') = 6/12
P(X' and Y') = 1/2
So the probability of having at least one good investment is 1/2
The final answer to part (A) is 1/2
---------------------------------------------
(B)
The events X and Y are independent so P(X and Y) = P(X)*P(Y)
P(X and Y) = P(X)*P(Y)
P(X and Y) = (1/3)*(1/4)
P(X and Y) = (1*1)/(3*4)
P(X and Y) = 1/12
The final answer to part B is 1/12
---------------------------------------------
(C)
Add the probabilities P(X) and P(Y). Then subtract off the probability found in part (B) to get the final answer
P(X or Y) = P(X) + P(Y) - P(X and Y)
Note: when I say P(X or Y) I mean you can only pick one event (X or Y). You cannot pick both. This is an exclusive 'or'.
So using that formula gives us this
P(X or Y) = P(X) + P(Y) - P(X and Y)
P(X or Y) = 1/3+1/4 - 1/12
P(X or Y) = 4/12+3/12 - 1/12
P(X or Y) = (4+3-1)/12
P(X or Y) = 6/12
P(X or Y) = 1/2
Answer to part C is 1/2
---------------------------------------------
(D)
"None of the investments would be a good stock" is the same as "Both investments are bad"
P(X' and Y') = P(X')*P(Y')
P(X' and Y') = (2/3)*(3/4)
P(X' and Y') = (2*3)/(3*4)
P(X' and Y') = 6/12
P(X' and Y') = 1/2