Question

There is a box with space for 16 items. Let A denote the number of things that
are type one and B the number of things that are
type two. Assume that A and B are independent
random variables. Assume that all possible (a,b)
pairs are equally likely.
I) How many possible pairs (a,b) are there?
II) Which event is more likely {A = 1} or {B = 0}?
Justify your answer.
III) Compute P(B=5) and P(A=10)
IV) If there are 10 things type 1, what is the
probability that there are 6 things type 2
in the box?
V) Compute P(B=5 | A=10)
VI) Compute P(B <= 5) and P(A <= 10)
VII) Compute P(B=5 | A <= 10)
VIII) Compute P(A+B=5) and P(A+B = 10)
IX) Compute P(A+B <= 5) and P(A+B <= 10)
X) Compute P(A+B <= 5 | A+B <= 10)
XI) Compute E(A+B)

Answer 1

Solution : There is a box with space for 16 items. Let A denote the number of things that
are type one and B the number of things that are type two. Assume that A and B are independent
random variables. Assume that all possible (a,b) pairs are equally likely.

Here we can see that,

A={1,2,3,..... ,16}

B={1,2,3,4......,16}

There can be 1 to 16 things of type A

and 1 to 16 things of type B

I) How many possible pairs (a,b) are there?

Its given that , all possible (a,b) pairs are equally likely.

let us consider ( 1 thing of A 15 things of B ), (2 thing of A 14 things of B ) and so on ...

There is space for 16 items does not mean that A + B =16 compulsarily

Hence we need to consider each and every possibility. We are assuming that box is not empty. Hence (0,0) pair will not come into picture.

(0,1),(0,2) ,(0,3),...(0,16) total= 16 pairs

(1 ,0), (1,1) , (1,2) ,.... (1,15) total = 16 pairs

(2,0) ,(2,1),(2,2),........(2,14) Total 15 pairs

(3,0),(3,1),.......(3,13) total 14 pairs up to

(15,0),(15,1) total 2 pairs

(16,0) 1 pair

Crriying on we get total pairs as

Total = 16+16+15+14+..........+2 +1

= 152

II) Which event is more likely {A = 1} or {B = 0}?
Justify your answer.

P(B=0) = 16 / 152

= 0.1052

P(A=1) = 16 / 152

=0.1052

Here {A=1} and { B= 0} are both equally likely.

III) Compute P(B=5) and P(A=10)

B can take 0 to 16 values

A can take 0 to 16 values

P(B=5 ) = 12 /152

= 0.0789

P(A=10) = 7 / 152

= 0.0460

{P(B=5) and P(A=10)} = P(B=5) * P(A=10)

= 0.0789 * 0.046

= 0.0036335

since they are independent random variable

IV) If there are 10 things type 1, what is the
probability that there are 6 things type 2
in the box?

V) Compute P(B=5 | A=10)

   P(B=5 | A=10) = {P(B=5) * P(A=10)} / P(A=10)

= P(B=5)

= 0.0789

VI) Compute P(B <= 5) and P(A <= 10)

We need to calculate P(B <= 5)

B can take 0,1,2,3,4,5

hence for B=0 there are 16 possibilities

B=1 there are 16 possibilities

B= 2 there are 15 possibilities

Carriying on we get ,

Hence P(B <= 5) = ( 16+16+15+14+13+12) / 152

= 0.565789

Similarly ,P(A <= 10)= (16+16+15+......+ 8+7) / 152

= 0.861842

P(B <= 5) and P(A <= 10) = P(B <= 5) * P(A <= 10)

= 0.565789 * 0.861842

= 0.487620

VII) Compute P(B=5 | A <= 10)

  hence for B=15 there are 12 possibilities

   P(B=5) = 12 / 152

   P(A <= 10)= (16+16+15+......+ 8+7) / 152

  P(B=5 | A <= 10) = (  P(B=5 ) * P( A <= 10) ) /   P( A <= 10)

= P(B=5 )

=  0.0789

VIII) Compute P(A+B=5) and P(A+B = 10)

= P(A+B=5) * P(A+B = 10)

= (6/ 152 )*( 11/152)

= 0.0028566

  

IX) Compute P(A+B <= 5) and P(A+B <= 10)

  Compute P(A+B <= 5)

Here A+ B can take values (1,2,3,.........16)

A+B <= 5 is  5+5+4+.. +1

number of pairs such that , A+B <= 5 are  20

and P(A+B <= 10)

(A+B <= 10 ) is 10+10+9+8+......+ 2+1

number of pairs such that ,(A+B <= 10 ) are 65

P(A+B <= 5) and P(A+B <= 10) = P(A+B <= 5) * P(A+B <= 10)

  = ( 20/ 152) *( 65/152)

    = 0.131578 * 0.4276

= 0.05626

X) Compute P(A+B <= 5 | A+B <= 10)

P(A+B <= 5 | A+B <= 10)=  P(A+B <= 5 ) *P( A+B <= 10)

= P(A+B <= 5 )

= 0.131578

XI) Compute E(A+B)

A+B	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16
N(A+B)	2	3	4	5	6	7	8	9	10	11	12	13	13	15	16	17
P(A+B)	0.013158	0.019737	0.026316	0.032895	0.039474	0.046053	0.052632	0.059211	0.065789	0.072368	0.078947	0.085526	0.085526	0.098684	0.105263	0.111842
	0.026316	0.059211	0.105263	0.164474	0.236842	0.322368	0.421053	0.532895	0.657895	0.796053	0.947368	1.111842	1.111842	1.480263	1.684211	1.901316

Summing up the product of P(A+B) and (A+B) in above table we get,
   Thus E( A+ B ) = ( A+B) * P( A+ B)

= 11.55921