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dangina [55]
3 years ago
12

Let’s fix some n e {1,2,3,...}. A biased coin with probability of head p is tossed n times, and the number of heads, N1, is coun

ted. The coin is then tossed Ni more times, and the number of heads, N2, is counted. Find the expected total number of heads, E[Ni + N2], generated by this process. The answer should depend on n and p only.
Mathematics
1 answer:
Yuki888 [10]3 years ago
7 0

Answer:

The answer is "np(1+p)".

Step-by-step explanation:

N_1 = Number of n tossed heads  

According to N_1, Y = amount of N_1 tossing heads

So, N_1~ Bin(n, p) and N_2 | N_1 ~ Bin(N_1, p)

E(N_1+N_2) & = E\Big[N_1+N_2\vert N_1\Big]\\ &

                   = E(N_1)+E\Big[N_2\vert N_1\Big]\\\\ & = E(N_1) + E(N_1\,p)\\\\ & = np + np.p \\\\ =np+np^2\\\\ =np(1+p)

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3 years ago
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finlep [7]

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Step-by-step explanation:

divide 54 by 9, and you get 6. this is because for every 5 sam gets bethan will get 4. if you add, that, you will get the total amount for 1 share. since there are 6 shares, you multiply that by the amount of eros sam and bethan gets each share.

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