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Suppose P(C) = 98%; P(D) = 88%; and P(C or D) = 33%.
Find P(C and D).
Can this really be a probability? If not, explain why not.

Respuesta :

MrRoyal

Answer:

[tex]P(C\ and\ D) = 153\%[/tex]

Step-by-step explanation:

Given

[tex]P(C) = 98\%[/tex]

[tex]P(D) = 88\%[/tex]

[tex]P(C\ or\ D) = 33\%.[/tex]

Required

[tex]P(C\ and\ D)[/tex]

In probability, the relationship between the given parameters and the required parameter is:

[tex]P(C\ and\ D) = P(C) + P(D) - P(C\ or\ D)[/tex]

Substitute values in the above formula

[tex]P(C\ and\ D) = 98\% + 88\% - 33\%[/tex]

[tex]P(C\ and\ D) = 153\%[/tex]

No, it can't be a probability

Because [tex]P(C\ and\ D) > 100\%[/tex]

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