MIOTP week 4 solutions

2)

BC²-BD²=31⁶

BC-BD=31²

BC+BD=31⁴

BC=31²(1+31²)/2

CosC=CD/BC

         =31/481

From this we can find the value of sinA and hence we can also find the value of CosA which is 480/481

m+n=961

4)A coin comes up head with probability p greater than 0 and tail with probability (1-p) greater than 0 suppose the probability of 4 heads and 5 tail is equal to 1/3 the probability of getting 5 head and 4 tail . Two person plays a game with two such coins the person who will first get two head will win the match what is the probability for the first person to win the match?

Answer: 16/23

Solution:

To solve this problem we need the knowledge of binomial probability

Our first aim is  to find the value of p using the given conditions 

Probability of getting 4 head 5 tail can be found out using Binomial probability

C(9,4)p⁴(1-p)⁵

Probability of getting 5 head 4 tail

Is C(9,5)p⁵(1-p)⁴

C(9,4)p⁴(1-p)⁵=1/3[C(9,5)p⁵(1-p)⁴]

3p⁴(1-p)⁵=p⁵(1-p)⁴

3(1-p)=p

p=3/4

Probability of getting two head is

P(2H)=3/4*3/4=9/16

Now there are infinite number of cases possible that is the person who is playing first can win in the first chance can win in the second chance can win in the third chance and so on

Case 1

The person wins in the 1st chance P=9/16

Case 2

The person who wins in the second chance means that the first person losses the first chance and  the second person also losses his first chance

and there are infinite cases

P(losing)=1-9/16=7/16

Total probability =7/16 × 7/16 × 9/16=(7/16)²9/16

Total probability

=9/16+(7/16)²9/16(7/16)⁴9/16..∞

=9/16(1+(7/16)²+(7/16)⁴...∞)

=9/16 * 256/207

=16/23

5) Solution Link