Question

Suppose you have a bag of 100 coins. Ninety of them are fair coins, which is P(H) = P(T) = 1/2. The other 10 coins are biased , they have tail (T) on both sides. Let X = {X1, X2, · · · , X100} be a random variable denoting the the total number of heads in the 100 coin flips.

a. How many possible values X can take? For example, if one tosses two fair coins, the number of total heads could be 0, 1, 2.

b. How many ways the value of X could be 1?

c. How many ways the value of X could be 2?

d. How many ways the value of X could be 25?

Answer 1

Ans -

Ninety of them are fair coins, which is P(H) = P(T) = 1/2

10 coins are biased , they have tail (T) on both sides

X_i denote the total number of heads in i^th draw.

(a)-

In 100 coins only 90's fair coins since only these 90 showing heads (remaining 10 are biased having tail on both the faces) in following manner

there is showing no head i.e X= 0

there is showing one head i.e X=1

there is showing two head i.e X =2 ............................................................

there is showing 90 head on 90 coin i.e X=90

therefore X ={ 0,1,2,3,......................90 }

.(b)-

Possible ways that value would be 1 = { there is 1 head on first coin are remaining coin having tail } or {there is 1 head at 2nd coin remaining having tail} ..............................or { there is 1 head at 90th coin  remaining having tail}

= 90 ways

.(C) - total ways the value of X could be 2 =  ⁹⁰ P ₂ ways = 8010 ways