Odds in favour of an event {\A} are 2 to 1 and odds in favour of {\A} \cup {\B} are 3 to 1 . Consistent with this information, the bound for {\P}( {\B}) is given by:

Correct Answer is : 1/12≤P( B )≤3/4

1 {/} 3 \leq {\P}( {\B}) \leq 1{/} 2

1{/} 12 \leq {\P}( B ) \leq 3{/} 4

0 \leq {\P}( {\B}) \leq 3 {/} 4

1{/} 6 \leq {\P}( {\B}) \leq 1{/} 3

NIMCET 2023 Question Paper with solutions

Question : 36 of 120

Marks: +1, -0

Odds in favour of an event

{\A}

are 2 to 1 and odds in favour of

{\A} ∪ {\B}

are 3 to 1 . Consistent with this information, the bound for

{\P}( {\B})

is given by:

Solution:

{\P}( {\A})\;=\;{2}/{3}

{\P}( {\A} ∪ {\B})\;=\;{3}/{4}

{\P}( {\A} ∪ {\B})\;= {\P}( {\A})+ {\P}( {\B})- {\P}( {\A} ∩ {\B})

\;{3}/{4}\;=\;{2}/{3}+ {\P}( {\B})- {\P}( {\A} ∩ {\B})

{\P}( {\A} ∩ {\B})\;= {\P}( {\B})-\;{1}/{12}

0 ≤ {\P}( {\A} ∩ {\B}) ≤ {\P}( {\A})

0 ≤ {\P}( {\B})-\;{1}/{12} ≤ \;{2}/{3}

\;{1}/{12} ≤ {\P}( {\B}) ≤ \;{2}/{3}+\;{1}/{12}

\;{1}/{12} ≤ {\P}( {\B}) ≤ \;{3}/{4} .

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