9. Suppose I have two coins: one fair and the otber biased that comes up heads w
ID: 3359219 • Letter: 9
Question
9. Suppose I have two coins: one fair and the otber biased that comes up heads with probability p 1/2. For each of the following, give the p.mf. Also state whether the distribution is binomial or not. If you thinkt is binomial, nclude any parameters (a) The number of hads among 10 tosses of the fair coin (b) The number of heads among 10 tosses of the biased coin. (c) The number of heads among 10 tosses of one of the coins that was selected at random from the two. (The same coin is used for all 10 tomes tosses (d) The number of heads among 10 tosses where before each toss one of the two coins is selected at random (e) The number of beads among 10 tosses where the finst 5 tosses used the fair coin and the rest of the tosses used the biased coin. (For this ne, instead of the p.m.f., just give the probability for 2 cases 10 heads and 0 heads.)Explanation / Answer
Question 9
There are two coins lets say C1 (fair) and C2 (biased)
For C1, let say number of heads are x1
P(X1) = 1/2 for X = Head
= 1/2 for X = tail
For C1 number of heads are x2
P(X1) = p for X = Head
= 1- p for X = tail
(a) Pr(Number of heads among 10 tosses of the fair coin) = Pr(X ; 10; 0.5) binomial distribution
N = 10 ; p = 0.5
(b) Pr(Number of heads among 10 tosses of the biased coin) = Pr(X ; 10 ; p) binomial distibution.
N = 10; p = p
(c) Here the probability of choosing one of the coins is 1/2
so if we choose one the same coin is used for all 10 tosses that will have a binomial distribution.
Pr( X number of heads) = 1/2 * BIN (X ; 10 ; 0.5) + 1/2 * BIN (X ; 10 ; p)
the distribution is not binomial.
(d) HEre Pr( Head in one toss) = Pr(Fair coin) * Pr(Head) + Pr(biased coin) * Pr(Head) = 1/4 + 1/2 * p = 1/4 + p/2
so Here Pr(X heads in 10 tosses) = BIN (X ; 10 ; 0.25 + 0.5p) so the given distribution is binomial.
(e) Pr(10 Heads) = Pr(all fair coins tosses will produce heads) + Pr(all biased coins tosses will produce heads)
Pr(10 Heads) = 0.55 + p5
Pr(0 Heads) = 0.55 + p5
so the given distribtion is not a binomial distribution.
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