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Re: The probability that student X will take history in the next [#permalink]
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It's 0.76, 4x0.19=0.76
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Re: The probability that student X will take history in the next [#permalink]
Could someone please clarify if correct answer is 0.76 rather 0.56?
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Re: The probability that student X will take history in the next [#permalink]
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Asmakan wrote:
The probability that student X will take history in the next semester is 0.75, the probability that at least one of students X and Y will take history in the next semester is 0.94, and whether or not one of students X and Y will take history in the next semester is independent of whether or not the other of these students will take history in the next semester. What is the probability that student Y will take history in the next semester?


Show: :: OA
0.56


may someone solves for me this question step by step ?

#s are changed due to rights



Let X = P(student X takes history)
Let Y = P(student Y takes history)

Given: The probability that student X will take history in the next semester is 0.75
So, P(X) = 0.75
Let P(Y) = k (since we aren't given this information)

Since events X and Y are independent, P(X and Y) = P(X) times P(Y)
= 0.75k

Also given: The probability that at least one of students X and Y will take history in the next semester is 0.94
In other words, P(X or Y) = 0.96

We're now ready to plug things into the OR probability formula...
P(X or Y) = P(X) + P(Y) - P(X and Y)
Substitute to get: 0.94 = 0.75 + k - 0.75k
Simplify the right side: 0.94 = 0.75 + 0.25k
Subtract 0.75 from both sides: 0.19 = 0.25k
Solve: k = 0.19/0.25 = 0.76

Answer: 0.76
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Re: The probability that student X will take history in the next [#permalink]
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Shimon wrote:
Could someone please clarify if correct answer is 0.76 rather 0.56?


I'm getting 0.76
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