Q12Section I (General (compulsory))12 of 84

The remaining probability $0.3$ splits $2:1$

A sample space consists of five simple events $E_1,\dots,E_5$ . If $P(E_1)=P(E_2)=0.15$ , $P(E_3)=0.4$ and $P(E_4)=2P(E_5)$ , then probability of $E_4$ and $E_5$ is:

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