Suppose I have two coins: one fair and the other biased that comes up heads with

Question

Suppose I have two coins: one fair and the other biased that comes up heads with probability p notequalto 1/2. For each of the following, write down the p.m.f., and decide whether the distribution is binomial or not. (a) The number of heads 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 tosses. The law of total probability might be useful here.) (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 heads among 10 tosses where the first 5 tosses used the fair coin and the rest of the tosses used the biased coin. (For this one, instead of the p.m.f., just give the probability for 2 cases: 10 heads and 0 heads.)

Explanation / Answer

Q.1 (a) PMF of a unbiased coin. Here X is the number of heads in 10 trials P(H) = 0.5

fX(x) = (10 C X) (0.5)10 [ Here 0 < = x < = 10)

(b) for biased coin.X is the number of heads in 10 trials P(H) = p

fX(x) = (10 C X) (p)X (1-p) 10-X [ Here 0 < = x < = 10) ]

the distributions are perfectly bionomial

(c) the probability of selecting one coin among these two coin = 0.5

so fX(x) = P(choosing fair coin) * P( X number of heads in 10 chances) +P(choosing biased coin) * P( X number of heads in 10 chances)

fX(x) = 0.5 [ (10 C X) (0.5)10 ] + 0.5 * (10 C X) (p)X (1-p) 10-X [ 0 <= X< = 10]

(d) Number of heads where before each toss one of the coins is selected at random.

so probability of getting Head in any of the toss = P(choosing fair coin) * 0.5 + P(choosing biased coin) * p

= 0.5 * 0.5 + 0.5 * p = (1 + 2p)/4

so in 10 tosses getting X number of heads

fX(x) = (10 C X) {(1+ 2p)/4)X {(3-2p)/4) 10-X [ where 0 <= X< = 10]

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