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Question : 39 of 160
Marks:
+1,
-0
Solution:
We have
P(E1∩E2)=P(E1)⋅P(‌) ∴‌‌P(E1∩E2)=‌⋅‌=‌ Now,
‌‌P(‌)=‌ ⇒‌‌‌=‌×‌ ⇒‌‌P(E2)=‌ Now,
P(‌)=‌ =‌=‌=‌×‌=‌
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