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Question

Mathematics Question on Probability

Consider two independent events EE and FF such thatP(E)=1/4,P(EF)=2/5 P(E)=1/4,P(E∪F)=2/5 and P(F)=aP(F)=a.Then,the value of aa is

A

1320\dfrac{13}{20}

B

120\dfrac{1}{20}

C

14\dfrac{1}{4}

D

15\dfrac{1}{5}

E

35\dfrac{3}{5}

Answer

15\dfrac{1}{5}

Explanation

Solution

Given Data

P(E)=14,P(EF)=25 P(E)=\dfrac{1}{4},P(E∪F)=\dfrac{2}{5}

We know that,

P(EF)=P(E)+P(F)P(EF)P(E ∪ F) = P(E) + P(F) - P(E ∩ F)

P(EF)=14+a25P(E ∩ F) = \dfrac{1}{4} + a - \dfrac{2}{5}---------(1)

Now, P(EF)=P(E)×P(F)P(E ∩ F) = P(E) × P(F)

P(EF)=(1/4)×aP(E ∩ F) = (1/4) × a----------(2)

Now, we can equate (1) and (2) to get the value of a'a’

14×a=14+a25\dfrac{1}{4} × a = \dfrac{1}{4} + a - \dfrac{2}{5}

14×aa=320⇒\dfrac{1}{4} × a -a = \dfrac{-3}{20}

3a4=320⇒\dfrac{-3a}{4}=\dfrac{-3}{20}

a=15⇒a=\dfrac{1}{5} (_Ans)