$P (E) = .24$ $P(E)=.24$ and $P (F) = .3$ $P(F)=.3$ . What is $P (E \cup F)$ $P(E uu F)$ ?

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Answer 1 · 2017-06-10T02:38:41

The probability of P or F occurring is .468

The formula is:

$P (A \cup B) = P (A) + P (B) - P (A \cap B)$ $P(AuuB)=P(A)+P(B)-P(AnnB)$

Basically you take the probability of $E + F$ $E +F$ minus the probability of both events happening.

In terms of $(E \cup F)$ $(EuuF)$ :

$P (E \cup F) = P (E) + P (F) - P (E \cap F)$ $P(EuuF)=P(E)+P(F)-P(EnnF)$

Plug in the values:

$P (E \cup F) = P (.24) + P (.3) - P (.24 \times .3)$ $P(EuuF)=P(.24)+P(.3)-P(.24xx.3)$

$P (E \cup F) = .24 + .3 - (.072)$ $P(EuuF)=.24+.3-(.072)$

$P (E \cup F) = .54 - .072$ $P(EuuF)=.54-.072$

$P (E \cup F) = .468$ $P(EuuF)=.468$

P(E)=.24P(E)=.24 and P(F)=.3P(F)=.3. What is P(E∪F)P(E uu F)?