The probability of an event of P(A∪B) is 49/128.
In this question,
Patricia will flip the coin 8 times.
The sample space is the flipping of a coin 8 times.
⇒ n(s) = 2⁸
⇒ n(s) = 256
Let A be the event that exactly 4 of the 8 flips are heads
⇒ n(A) = [tex]8C_4[/tex] [∵ [tex]nC_r=\frac{n!}{r!(n-r!)}[/tex]]
⇒ [tex]\frac{8!}{4!4!}[/tex] = 70
Then, P(a) = [tex]\frac{n(a)}{n(s)}[/tex]
⇒ [tex]\frac{70}{256}[/tex]
Let B be the event that the last two flips are tails
⇒ n(B) = [tex]8C_2[/tex]
⇒ [tex]\frac{8!}{2!6!}[/tex] = 28
Then, P(a) = [tex]\frac{n(b)}{n(s)}[/tex]
⇒ [tex]\frac{28}{256}[/tex]
Now, P(A∪B) = P(A) + P(B)
⇒ P(A∪B) = [tex]\frac{70}{256} + \frac{28}{256}[/tex]
⇒ P(A∪B) = [tex]\frac{70+28}{256}[/tex]
⇒ P(A∪B) = [tex]\frac{98}{256}[/tex]
⇒ P(A∪B) = [tex]\frac{49}{128}[/tex]
Hence we can conclude that the probability of an event of P(A∪B) is 49/128.
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