A biased coin shows heads with probability 0.3. If it is tossed 5 times, what is the probability of getting no more than 2 heads?

We are given:

• A biased coin has P(H) = 0.3, so P(T) = 0.7
• The coin is tossed 5 times
• We need to find the probability of getting no more than 2 heads, i.e., 0, 1, or 2 heads

This is a binomial probability problem. Use binomial formula The binomial probability formula is: P(X = k) = C(n, k) × (p)^k × (1 – p)^n-k Here:

• n = 5 (number of trials)
• p = 0.3 (probability of head)
• k = number of heads (we’ll calculate for 0, 1, and 2)

Calculate individual probabilities P(0 heads) = C(5, 0) × (0.3)^0 × (0.7)^5 = 1 × 1 × 0.16807 = 0.16807 P(1 head) = C(5, 1) × (0.3)^1 × (0.7)^4 = 5 × 0.3 × 0.2401 = 5 × 0.07203 = 0.36015 P(2 heads) = C(5, 2) × (0.3)^2 × (0.7)^3 = 10 × 0.09 × 0.343 = 10 × 0.03087 = 0.3087 Add the probabilities P(X ≤ 2) = P(0) + P(1) + P(2) = 0.16807 + 0.36015 + 0.3087 = 0.83692 Rounding to 3 decimal places: 0.837