Question

$A\mathrm{pair}\mathrm{of}\mathrm{dice}\mathrm{is}\mathrm{rolled}\mathrm{and}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{the}\mathrm{nos}.\mathrm{that}\mathrm{comes}\mathrm{up}\mathrm{is}\mathrm{noted}\mathrm{down},\mathrm{then}\mathrm{consider}\mathrm{the}\mathrm{following}\mathrm{events}-A=\mathrm{sum}\mathrm{of}\mathrm{nos}.\mathrm{is}\mathrm{odd}B=\mathrm{sum}\mathrm{of}\mathrm{nos}.\mathrm{is}\mathrm{even}C=\mathrm{sum}\mathrm{of}\mathrm{nos}.\mathrm{is}\mathrm{more}\mathrm{than}11D=\mathrm{sum}\mathrm{of}\mathrm{nos}.\mathrm{is}\mathrm{less}\mathrm{than}4E=\mathrm{sum}\mathrm{of}\mathrm{nos}.\mathrm{is}\mathrm{prime}\mathrm{no}.$

• A. Exclusive events
• B. Exhaustive events
• C. Simple event
• D. Equally likely event

Answer 1

The sample space S = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
A = Sum of nos. is odd
= {3, 5, 7, 9, 11}
B = Sum of nos. is even
= {2, 4, 6, 8, 10, 12}
C = Sum of nos. is more than 11
= {12}
D = Sum of nos is less than 4
={2, 3}
E = Sum of nos. is prime no.
={2, 3, 5, 7, 11}

$A\cup B$ will contain all the elements $A\cup B$= {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
so, A and B is exhaustive events

$\mathrm{C\; \&}D$ does not have any common element so they are exclusive events

$B\cap C=\left\{12\right\}$ has only one element so it is a simple event

$A\&E$ = both have equal no of elements so they are equally likely events