find the distribution function of the total number of heads obtained in four tosses of a biased coin About the author Aaliyah

Step-by-step explanation: When a coin is tossed we have S={H,T} P (getting a head ) =12 and P (not getting a head) =(1−12)=12 Let X be the random variable denoting the number of heads In 4 trials we may get 0 or 1 or 2 or 3 or 4 heads So. X may assume the values 0,1,2,3,4 . P(X=0)=.4C0.(12)0.(12)(4−0)=116 P(X=1)=.4C1.(12)1.(12)(4−1)=14 P(X=2)=.4C2.(12)2.(12)(4−2)=38 P(X=3)=.4C3.(12)3.(12)(4−3)=14. P(X=4)=.4C4.(12)4.(12)(4−4)=116 Hence the required probability distribution is given by (X:,0,1,2,3,4P(X):,116,14,38,14,116) Reply

Step-by-step explanation:When a coin is tossed we have S={H,T}

P (getting a head ) =12 and

P (not getting a head) =(1−12)=12

Let X be the random variable denoting the number of heads

In 4 trials we may get 0 or 1 or 2 or 3 or 4 heads

So. X may assume the values 0,1,2,3,4 .

P(X=0)=.4C0.(12)0.(12)(4−0)=116

P(X=1)=.4C1.(12)1.(12)(4−1)=14

P(X=2)=.4C2.(12)2.(12)(4−2)=38

P(X=3)=.4C3.(12)3.(12)(4−3)=14.

P(X=4)=.4C4.(12)4.(12)(4−4)=116

Hence the required probability distribution is given by

(X:,0,1,2,3,4P(X):,116,14,38,14,116)