Solution
Given:In this question, we are given
• In a box, 20 bulbs are kept and some of them are defective.
• 2 bulbs are randomly selected from the box.
To find:We need to determine
• If the probability that both selected bulbs are defective is less than 0.4 or not.
Approach and Working:Let us assume that among 20 bulbs, N number of bulbs are defective.
• Therefore, the probability that 2 selected balls will be defective = \(^NC_2/^{20}C_2 = \frac{N(N – 1)}{2} * \frac{2}{19 * 20} = \frac{N(N – 1)}{19 * 20}\)
Now, if the probability is to be less than 0.4, then we can say
• \(\frac{N(N – 1)}{19 * 20} < 0.4\)
Or, N(N – 1) < 152
As N is an integer, N must be less than 13.
Hence, we need to know whether N < 13 or not.
Analysing Statement 1As per the information given in statement 1, more than 50% of the bulbs are defective.
• Therefore, we can say N > 10
• However, we cannot conclude whether N < 13 or not.
Hence, statement 1 is not sufficient to answer the question.
Analysing Statement 2As per the information given in statement 2, less than 60% of the bulbs are defective.
• Therefore, we can say that N < 12
• As N < 12, we can definitely say N < 13.
Hence, statement 2 is sufficient to answer the question.
Hence, the correct answer is option B.
Answer: B