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Question : 37 of 160
Marks:
+1,
-0
Solution:
‌‌ Given, ‌P(B1)=0.25,P(B2)=0.30‌P(B3)=0.45,P(‌)=0.05‌P(‌)=0.04,P(‌)=0.03∴P(‌)=‌| P()P(B2) |
| P(‌)P(B2)+P(‌)P(B3) |
‌‌+P(‌)P(B1)0.04×0.30‌=‌| 0.03×0.45+0.05×0.25 |
| 0.04×0.30+0.03 |
‌=‌‌=‌=‌
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