TIFR CSE 2015 | Part A | Question: 1

Question

Consider a $6$-sided die with all sides not necessarily equally likely such that probability of an even number is $P (\left \{2, 4, 6  \right \}) =\dfrac{1}{2}$, probability of a multiple of  $3$ is $P (\left \{3, 6  \right \}) = 1/3$ and probability of $1$ is $P(\left \{ 1 \right \}) = \dfrac{1}{6}$. Given the above conditions, choose the strongest (most stringent) condition of the following that must always hold about $P(\left \{ 5 \right \})$, the probability of $5$.

• A. $P(\left \{5  \right \}) =\dfrac{1}{6}$
• B. $P(\left \{ 5 \right \}) \geq \dfrac{1}{6}$
• C. $P(\left \{5  \right \}) \leq  \dfrac{1}{6}$
• D. $P(\left \{ 5 \right \}) \leq  \dfrac{1}{3}$
• E. None of the above

Comments

'option d is correct'

Answer 1

P(2,4,6) =1/2   , P(3,6)= 1/3 , P(1)= 1/6

So,  P(2,4,6) +P(3,6) - P(6) + P(1) + P(5) =1

    or,   1/2 + 1/3  - P(6) + 1/6 + P(5) =1

     or, 1 - P(6) + P(5) =1

    or, P(5) =P(6)

Now we know, P(3,6)=1/3

or, P(3) + P(6) = 1/3

or, P(3) + P(5) =1/3  [ since P(5)=P(6)]

or, P(5) =1/3 - P(3)

Since P(3)>=0,  P(5)<=1/3