GO Classes 2024 | Weekly Quiz 8 | Conditional Probability | Question: 12

Answer 1

$p(M\cup B)$ will be the maximum when $M$ and $B$ are mutually exclusive. In this case,  $p(M\cup B)$ = $p(M)$ + $p(B)$ = 0.9

$p(M\cup B)$ will be the minimum when one event is a proper subset of another.  Here $M$ has to be a proper subset of $B$ because the number of elements in $M$ is less than $B$. In this case, $p(M\cup B)$ = $p(B)$ = 0.5

$\therefore$  0.5 <= p <= 0.9

Answer 2

The easy way to answer this is that $A \cup B$ has a minumum of 25 members (when all males are brown-eyed) and a maximum of 45 members (when no males have brown-eyes). So, the probability ranges from .5 to .9 Thinking about it in terms of the inclusion-exclusion principle we have
$$
P(M \cup B)=P(M)+P(B)-P(M \cap B)=.9-P(M \cap B) .
$$
So the maximum possible value of $P(M \cup B)$ happens if $\mathrm{M}$ and $B$ are disjoint, so $P(M \cap B)=0$. The minimum happens when $M \subset B$, so $P(M \cap B)=P(M)=.4$

Answer 3

M = (M ∩ B ) U (M ∩ B’ ) and (M ∩ B ), (M ∩ B’ ) are disjoint from the venn diagram so p(M) = p(M ∩ B ) + p(M ∩ B’ ) it means

p(M ∩ B ) =p(M) - p(M ∩ B’ )  which means p(M ∩ B ) <=p(M)

 

 so p(M ∩ B ) <= p(M)

p( M ∩ B) <=p(B)

so p(M∩B) <= min(p(M),p(B)) so p(M∩B) <= min(0.4,0.5) so p(M∩B)<=0.4 if M and B are not disjoint

so if M and B are not disjoint then p(MUB) = 0.4 +0.5 -0.4 = 0.5 and

if M and B are disjoint then p(MUB) = 0.4+0.5 =0.9

so 0.5<=p<= 0.9 is range of p which is option D

Sachin Mittal 1 sir can u please confirm is my method correct ?