(a) Two cards are drawn at random and one by one with replacement from a well-shuffled pack of 52 playing cards. Find the probability distribution of the number of aces. Also, find its mean and variance. OR (b) It is given that 2% of the screws manufactured by a company are defective. Use Poisson distribution to find the probability that a packet of 100 screws contains (i) no defective screw, (ii) one defective screw.
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Step-by-Step Solution
Step 1: Identify parameters for Poisson distribution
We are given that the number of screws in a packet is n=100 and the probability of a screw being defective is p=2%=0.02. For a Poisson distribution, we need to calculate the mean λ.
Step 2: Calculate the mean (λ)
The mean λ for a Poisson distribution, when approximating a binomial distribution, is given by the product of the number of trials n and the probability of success p. In this case, λ=100×0.02=2.
Step 3: State the Poisson probability formula
The probability mass function for a Poisson distribution is given by P(X=k)=ke−λλk!, where k is the number of occurrences, λ is the average rate of occurrence, and e is Euler's number (approximately 2.71828).
Step 4: Calculate probability of no defective screws (k=0)
To find the probability of no defective screws, we substitute k=0 and λ=2 into the Poisson probability formula. Since 20=1 and 0!=1, the probability simplifies to e−2, which is approximately 0.1353.
Step 5: Calculate probability of one defective screw (k=1)
To find the probability of one defective screw, we substitute k=1 and λ=2 into the Poisson probability formula. This gives us 2e−2, which is approximately 2×0.1353=0.2706.