# TIFR2015-A-1

993 views

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 $5$.

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

edited
1
'option d is correct'

$P\{3,5\} = 1 - P\{2,4,6\} - P\{1\} =\dfrac{1}{2} -\dfrac{1}{6} =\dfrac{1}{3}$

Can $P{3} = 0$ ? then $P\{6\} =\dfrac{1}{3}$ and $P\{2,4\} =\dfrac{1}{2} -\dfrac{1}{3}=\dfrac{1}{6}.$

And

$P\{5\} =\dfrac{1}{3}.$ Possible.

So, option D.

selected by
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.
0

yes, if P(6)=0 ,

=> P(3) = 1/3

=> P(5) = 0.

0<= P(5) <=1/3

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 $P(3)+P(6)$ = $\frac{1}{3}$
$P(6)$ = $\frac{1}{3} - P(3)$
0
So how 1/3?
5

P(5) = P(6) = 1/3 - P(3)

min value of P(5) : when P(3) has max value 1/3 .so min value of P(5) =1/3 -1/3 =0

max value of P(5):when P(3)​​​​​​​ has min value 0 .so max value of P(5) =1/3 -​​​​​​​0 =1/3

So,  0≤ P(5)≤1/3  or we can say P(5) ≤ 1/3.

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

0
nice

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

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

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

=> 1/2 + 1/3 + 1/6 + P(5) = 1 + P(6)

=> P(5) = P(6)

Also, P(6)<=1/6 and P(6)<=1/3. In the worst case, P(5)=P(6)<=1/3
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

## Related questions

1
565 views
Consider two independent and identically distributed random variables $X$ and $Y$ uniformly distributed in $[0, 1]$. For $\alpha \in \left[0, 1\right]$, the probability that $\alpha$ max $(X, Y) < XY$ is $1/ (2\alpha)$ exp $(1 - \alpha)$ $1 - \alpha$ $(1 - \alpha)^{2}$ $1 - \alpha^{2}$
2
2.2k views
Ram has a fair coin, i.e., a toss of the coin results in either head or tail and each event happens with probability exactly half $(1/2)$. He repeatedly tosses the coin until he gets heads in two consecutive tosses. The expected number of coin tosses that Ram does is. $2$ $4$ $6$ $8$ None of the above.
3
415 views
Let $A$ and $B$ be non-empty disjoint sets of real numbers. Suppose that the average of the numbers in the first set is $\mu_{A}$ and the average of the numbers in the second set is $\mu_{B}$; let the corresponding variances be $v_{A}$ and $v_{B}$ respectively. If the average of the elements in $A \cup B$ ... $p.v_{A}+ (1 - p). v_{B} + (\mu_{A}- \mu_{B})^{2}$
Consider the following $3 \times 3$ matrices. $M_{1}=\begin{pmatrix} 0&1&1 \\ 1&0&1 \\ 1&1&0 \end{pmatrix}$ $M_{2}=\begin{pmatrix} 1&0&1 \\ 0&0&0 \\ 1&0&1 \end{pmatrix}$ How may $0-1$ column vectors of the form $X$ ... $2$ means all operations are done modulo $2$, i.e, $3 = 1$ (modulo $2$), $4 = 0$ (modulo $2$)). None Two Three Four Eight