10 votes

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 $5$.

- $P(\left \{5 \right \}) =\dfrac{1}{6}$
- $P(\left \{ 5 \right \}) \geq \dfrac{1}{6}$
- $P(\left \{5 \right \}) \leq \dfrac{1}{6}$
- $P(\left \{ 5 \right \}) \leq \dfrac{1}{3}$
- $\text{None of the above.}$

10 votes

Best answer

1

over all probability =P(1)+P(2)+P(3)+P(4)+P(5)+P(6)=1

Now P(1)+P(3)+P(2)+P(4)+P(6)=1/6+1/2+1/3=1.

So it is possible P(5)=0? .Please correct me if i am wrong.

Now P(1)+P(3)+P(2)+P(4)+P(6)=1/6+1/2+1/3=1.

So it is possible P(5)=0? .Please correct me if i am wrong.

20

$P(A \cup B)$ = $P(A)+P(B)-P(A \cap B)$ (in this question, all intersection probabilities are zero)

$P(2 \cup 4 \cup 6) $ = $P(2)+P(4)+P(6)$ = $\frac{1}{2}$ $(1)$

$P(3 \cup6) $ = $P(3)+P(6)$ = $\frac{1}{3}$ $(2)$

$P(1)$ = $\frac{1}{6}$ $(3)$

$P(1)+P(2)+P(3)+P(4)+P(5)+P(6)$ = $1$ $(4)$

Adding equation $1$, $2$ and $3$

$P(1)+P(2)+P(3)+P(4)+2P(6)$ = $\frac{1}{2}+\frac{1}{3}+\frac{1}{6}$

$P(1)+P(2)+P(3)+P(4)+2P(6)$ = $1$

using equation $4$ and above equation, we can say that $P(5)=P(6)$

and it is given that $P(3)+P(6)$ = $\frac{1}{3}$

$P(6)$ = $\frac{1}{3} - P(3)$

$P(2 \cup 4 \cup 6) $ = $P(2)+P(4)+P(6)$ = $\frac{1}{2}$ $(1)$

$P(3 \cup6) $ = $P(3)+P(6)$ = $\frac{1}{3}$ $(2)$

$P(1)$ = $\frac{1}{6}$ $(3)$

$P(1)+P(2)+P(3)+P(4)+P(5)+P(6)$ = $1$ $(4)$

Adding equation $1$, $2$ and $3$

$P(1)+P(2)+P(3)+P(4)+2P(6)$ = $\frac{1}{2}+\frac{1}{3}+\frac{1}{6}$

$P(1)+P(2)+P(3)+P(4)+2P(6)$ = $1$

using equation $4$ and above equation, we can say that $P(5)=P(6)$

and it is given that $P(3)+P(6)$ = $\frac{1}{3}$

$P(6)$ = $\frac{1}{3} - P(3)$

11 votes

Given,

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

P(3,6)= 1/3

P(1)= 1/6

So,

P(1,3,5)=1/2

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

= 1/2 - 1 /6

= 1/3

P(3,5)=1/3

For P(5) to have the maximum probablity, P(3) should be 0

P(5)<=1/3

6 votes

D is the correct answer. so lets find out how. first thing keep in mind is that the chances are not equally likely. so probability can be go like 99:1 also . so

the given data is P(2,4,6) =1/2

so P(1,3,5)= 1/2. here it does not mean P(1)=P(3). but we an surely say the sum will be 1/2.

so P(3,6)= 1/3 again sum can be 1/6 . individual probability may not be equal . so lets find the maximum probability that can exist for 5.

the maximum probability for 5 will be when there will be minimum probaility fo 1 and 3. and P(3,6) =1/3 = 0.33333. so i considered that the possible minimum probability of 3 will be 0.01. and 6 probability will be 0.32. again we get a sum of 0.333.

now one probability is already given . which is 1/6= 0.16.

minimum probability for both 3 and 1 = 0.16+0.01=0.17.

so the maximum probability for 5 will be 0.5 i.e (1/2)-0.17.

0.5-0.17= 0.33. which is equal to 1/3. ( this is the maximum probability of 5) . so we can say P({5})≤1/3

P({5})≤1/3P({5})≤1/3