In four tosses of a coin, let X be the number of heads, Tabulate the 16 possible outcomes with the corresponding value of X. Calculate the expected value of X.
Answers
Answered by
1
If XX is the number of heads in 44 tosses of a coin, we can have X=0X=0, X=1X=1, X=2X=2, X=3X=3, or X=4X=4. Assuming that the tosses are independent, XX will have a binomial distribution, Bin(4,12)Bin(4,12). This means
P(X=0)=(40)(12)0(12)4−0=116,P(X=1)=(41)(12)1(12)4−1=14,P(X=2)=(42)(12)2(12)4−2=38,P(X=3)=(43)(12)3(12)4−3=14, andP(X=4)=(44)(12)4(12)4−4=116
P(X=0)=(40)(12)0(12)4−0=116,P(X=1)=(41)(12)1(12)4−1=14,P(X=2)=(42)(12)2(12)4−2=38,P(X=3)=(43)(12)3(12)4−3=14, andP(X=4)=(44)(12)4(12)4−4=116
It follows that
E((2X+1)2)=(12)(116)+(32)(14)+(52)(38)+(72)(14)+(92)(116)=29
E((2X+1)2)=(12)(116)+(32)(14)+(52)(38)+(72)(14)+(92)(116)=29
Similar questions