1. **Problem Statement:** From a lot of 30 bulbs including 6 defective bulbs, a sample of 2 bulbs is drawn at random one by one with replacement. Find the probability distribution of the number of defective bulbs and the mean number of defective bulbs. 2. **Understanding the problem:** Since the sampling is with replacement, the probability of drawing a defective bulb remains constant for each draw. 3. **Define random variable:** Let $X$ be the number of defective bulbs in the 2 draws. Possible values of $X$ are 0, 1, or 2. 4. **Calculate probabilities:** - Probability of defective bulb in one draw: $p = \frac{6}{30} = 0.2$ - Probability of non-defective bulb: $q = 1 - p = 0.8$ 5. **Probability distribution:** Since draws are independent, - $P(X=0) = q \times q = 0.8 \times 0.8 = 0.64$ - $P(X=1) = (p \times q) + (q \times p) = 2 \times 0.2 \times 0.8 = 0.32$ - $P(X=2) = p \times p = 0.2 \times 0.2 = 0.04$ 6. **Mean (Expected value) of $X$:** $$E(X) = 0 \times 0.64 + 1 \times 0.32 + 2 \times 0.04 = 0 + 0.32 + 0.08 = 0.4$$ **Final answer:** - Probability distribution: $P(X=0)=0.64$, $P(X=1)=0.32$, $P(X=2)=0.04$ - Mean number of defective bulbs: $0.4$