Question

1.1. Suppose that a fair coin is flipped 6 times in sequence and let X be the number of "heads" that show up. Draw Pascal's triangle down to the sixth row (recall that the zeroth row consists of a single 1) and use your table to compute the probabilities P(X k) for k 0,1,2,3, 4,5,6

Answer 1

Pascal’s Triangle:

Pascal’s Triangle for Combinations:

Row 0: 1

Row 1: 1C0 and 1C1

Row 2: 2C0, 2C1 and 2C2

Row 3: 3C0, 3C1, 3C2 and 3C3

Row 4: 4C0, 4C1, 4C2, 4C3 and 4C4

Row 5: 5C0, 5C1, 5C2, 5C3, 5C4 and 5C5

Row 6: 6C0, 6C1, 6C2, 6C3, 6C4, 6C5 and 6C6

n = 6, p = 1/2, q = 1 – p = 1/2

P(k) = nCk p^k q^(n – k)

P(0) = 6C0 (1/2)^0 (1/2)^(6 – 0) = 1 * (1/2)^0 * (1/2)^6 = 1/64

P(1) = 6C1 (1/2)^1 (1/2)^(6 – 1) = 6 * (1/2) * (1/2)^5 = 3/32

P(2) = 6C2 (1/2)^2 (1/2)^(6 – 2) = 15 * (1/2)^2 * (1/2)^4 = 15/64

P(3) = 6C3 (1/2)^3 (1/2)^(6 – 3) = 20 * (1/2)^3 * (1/2)^3 = 5/16

P(4) = 6C4 (1/2)^4 (1/2)^(6 – 4) = 15 * (1/2)^4 * (1/2)^2 = 15/64

P(5) = 6C5 (1/2)^5 (1/2)^(6 – 5) = 6 * (1/2)^5 * (1/2) = 3/32

P(6) = 6C6 (1/2)^6 (1/2)^(6 – 6) = 1 * (1/2)^6 * (1/2)^0 = 1/64