Probability Test - 2

Let X denote same number of heads Both will get.

So X can take values 

X=0 ( both get 0 number of head)

X =1 ( both get 1 head)

X = 2 ( both get 2 heads)

X = 3 ( both get 3 heads)

X = 4 ( both get 4 heads)

So , P(X = 0 ) =Probabilty of getting no heads= 1/16 × 1/16 = 1/256      [by multiplication theorem  P(X_i∩∩ X_j) = P(X_i).P(X_j) ]

P(X=1) =  Probabilty of getting 1 head = 4/16 × 4/16 = 16/256

P(X =2) = Probabilty of getting 2 heads =  6/16 × 6/16 = 36/256

P ( X = 3) =Probabilty of getting 3 heads=4/16 × 4/16 = 16/256

P( X= 4) =Probabilty of getting 4 heads = 1/16 × 1/16 = 1/256

Therefore P ( both get same number of head in a throw of 4 coins each )

= P(X=0) + P(X=1) + P(X=2) + P(X =3) + P(X =4)

= 1/256 + 16/256 + 36/256 + 16/256 + 1/256

= 70/256 = 35/128

When three dice  are rolled, total number of outcomes =n(S)=n=6³ =216

It is given that the numbers on all the three dice are identical

Hence the favourable outcomes are (1,1,1),(2,2,2),(3,3,3),(4,4,4),(5,5,5),(6,6,6)

Hence no. of  favourable outcomes=m=6

Hence required probability=m/n=6/216=1/36

Let X denote getting an even number on each dice on throwing twice.

By only one way it can be done by having 2 in both dice.

Since 2 occupy two faces of dice.

So P(getting 2 in 1st dice) = 2/6

Also P ( getting 2 in 2nd dice) = 2/6

So P ( getting 2 in both dice dice) = 2/6 × 2/6 = 1/9 [from multiplication theorem of probabilty P(A ∩ B) = P(A)xP(B) for any two non empty sets A, B]

Sample Space is

S = { HH, HT, T1, T2, T3, T4, T5, T6 }

so number of outcomes in sample space is 8

P(A wins)=1/9+8/9*5/6*1/9+8/9*5/6*8/9*5/6*1/9+...........=(1/9)/(1-8/9*5/6)=3/7

Total Number of 2nd order determinants possible with entries only 0 and 1 is ^{​​​​​​}2⁴ = 16              

Out of these 16 different determinants only three determinants will have value in positive.

So probability of getting 3 favourable item out of 16 is 3/16.