Abu visits his local hardware store to buy six light bulbs - AQA - A-Level Maths Pure - Question 15 - 2018 - Paper 3

To find the probability that all six bulbs are faulty, we use the binomial probability formula:

P(X = k) = inom{n}{k} p^k (1-p)^{n-k}

Substituting for 6 faulty bulbs: P(X = 6) = inom{6}{6} (0.15)^6 (0.85)^{0} = 1 imes (0.15)^6 imes 1 = 0.0001134. Therefore, the probability is approximately 0.000114.

To find the probability that at least two bulbs are faulty, we can use the complementary approach: $P(X ext{ at least } 2) = 1 - P(X < 2) = 1 - (P(X = 0) + P(X = 1))$.

Calculating:

• For no faulty bulbs: P(X = 0) = inom{6}{0} (0.15)^0 (0.85)^6 = 1 imes 1 imes (0.85)^6 = 0.5277.
• For one faulty bulb: P(X = 1) = inom{6}{1} (0.15)^1 (0.85)^5 = 6 imes (0.15) imes (0.85)^5 = 0.3861.

Then, $P(X < 2) = P(X = 0) + P(X = 1) = 0.5277 + 0.3861 = 0.9138$.

So, $P(X ext{ at least } 2) = 1 - 0.9138 = 0.0862.$

The mean of a binomial distribution can be found using the formula: $ext{Mean} = n imes p$ Where n is the number of trials (6) and p is the probability of success (0.15).

Thus, the mean is: $ext{Mean} = 6 imes 0.15 = 0.9$.