Multiple choice

An unbiased normal coin is tossed n times. Let $E_1$: event that both heads and tails are present in n tosses. $E_2$:event that the coin shows up heads at most once. The value of n for which $E_1$ and $E_2$ are independent is

  1. 1

  2. 2

  3. 3

  4. 4
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

E1: At least one H and one T. P(E1) = 1 - (P(all H) + P(all T)) = 1 - (1/2^n + 1/2^n) = 1 - 2/2^n = 1 - 1/2^(n-1). E2: At most one H. P(E2) = P(0 H) + P(1 H) = (1 + n)/2^n. Independence requires P(E1 intersect E2) = P(E1) * P(E2). For n=3, P(E1)=6/8=3/4, P(E2)=4/8=1/2. P(E1 intersect E2) = P(exactly 1 H) = 3/8. Since 3/4 * 1/2 = 3/8, n=3 is correct.

AI explanation

To test for independence, we use the fundamental identity P(E1 intersection E2) = P(E1) * P(E2). We calculate P(E1) = 1 - 1/2^n, P(E2) = (n+1)/2^n, and P(E1 intersection E2) = n/2^n. Setting n/2^n = ((2^n - 1)/2^n) * ((n+1)/2^n) and testing n = 3 yields 3/8 = (7/8) * (4/8), proving both sides equal 7/16.

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