612 views
Let $A, B$ and $C$ be independent events which occur with probabilities $0.8, 0.5$ and $0.3$ respectively. The probability of occurrence of at least one of the event is _____
| 612 views

$P(A) = 0.8$ $\implies$ $P(A') =$ $1 - 0.8 = 0.2$

$P(B) = 0.5$ $\implies$ $P(B')$ = $1 - 0.5 = 0.5$

$P(C) = 0.3$ $\implies$ $P(C')$ = $1 - 0.3 = 0.7$

$P$(No event will occur): =$0.2 * 0.5 * 0.7 = 0.07$

$P$(at least 1 event will occur): $1 - 0.07 = 0 .93$
by Boss (42.6k points)
edited
0

or

We can calculate like this

$0.8\times 0.5\times 0.7+0.2\times 0.5\times 0.7+0.2\times 0.5\times 0.3+0.8\times 0.3\times 0.5+0.8\times 0.5\times 0.7+0.5\times 0.3\times 0.2+0.8\times 0.3\times 0.5=0.93$

Inclusion-Exclusion principle is not working here

P(at least A or B or C )= P(A u B u c)

= 0.8+0.5+0.3-(0.8*0.5) -(0.5 *0.3) -(0.8*0.3)+(0.8*0.5*0.3)

=1.6-0.67

=0.93
by Active (4.2k points)
0
This is right but above method is fast.

1
2