S5 A bag contains twenty counters ,numbered 1.2、 A counter is selected at random from the bag its number IS noted , and it is then returned to the ba (i)If this operation is carried out 100 times find, 2 significant figures , the probability that the num selected at least A times. (ii)If the operation is carried out 3 times find th probability that the largest number drawn is n . W 1leqslant nleqslant 20 . Find the expected value of n to 3 sig figures.

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S5 A bag contains twenty counters ,numbered 1.2、 A counter is selected at random from the bag its number IS noted , and it is then returned to the ba (i)If this operation is carried out 100 times find, 2 significant figures , the probability that the num selected at least A times. (ii)If the operation is carried out 3 times find th probability that the largest number drawn is n . W 1leqslant nleqslant 20 . Find the expected value of n to 3 sig figures.

Answer 1

(i) The probability that the number is selected at least 4 times in 100 trials is approximately \(0.98\) (to 2 significant figures). (ii) The expected value of the largest number drawn in 3 trials is approximately \(14.1\) (to 3 significant figures).

Answer 2

## Step 1: Identify the total number of counters and the probability of selecting any one counter. There are 20 counters, each numbered from 1 to 20. The probability of selecting any one counter is: \[P(\text{selecting any specific counter}) = \frac{1}{20}\] ## Step 2: Calculate the probability of selecting a counter at least 4 times in 100 trials. We use the binomial distribution formula \(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\), where \(n\) is the number of trials, \(k\) is the number of successes, and \(p\) is the probability of success in a single trial. Here, \(n = 100\), \(k \geq 4\), and \(p = \frac{1}{20}\). The probability of selecting a counter at least 4 times is: \[P(X \geq 4) = 1 - P(X < 4)\] \[P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)\] Calculate each term: \[P(X = 0) = \binom{100}{0} \left(\frac{1}{20}\right)^0 \left(1 - \frac{1}{20}\right)^{100}\] \[P(X = 1) = \binom{100}{1} \left(\frac{1}{20}\right)^1 \left(1 - \frac{1}{20}\right)^{99}\] \[P(X = 2) = \binom{100}{2} \left(\frac{1}{20}\right)^2 \left(1 - \frac{1}{20}\right)^{98}\] \[P(X = 3) = \binom{100}{3} \left(\frac{1}{20}\right)^3 \left(1 - \frac{1}{20}\right)^{97}\] Sum these probabilities and subtract from 1 to get \(P(X \geq 4)\). ## Step 3: Calculate the probability of the largest number drawn being \(n\) in 3 trials. The probability of the largest number drawn being \(n\) is given by: \[P(\text{largest number} = n) = P(\text{all numbers} \leq n) - P(\text{all numbers} \leq n-1)\] For \(n\) trials, the probability that all numbers are \(\leq n\) is: \[P(\text{all numbers} \leq n) = \left(\frac{n}{20}\right)^3\] And for \(n-1\): \[P(\text{all numbers} \leq n-1) = \left(\frac{n-1}{20}\right)^3\] Thus: \[P(\text{largest number} = n) = \left(\frac{n}{20}\right)^3 - \left(\frac{n-1}{20}\right)^3\] ## Step 4: Calculate the expected value of \(n\). The expected value \(E(n)\) is given by: \[E(n) = \sum_{n=1}^{20} n \cdot P(\text{largest number} = n)\] Substitute the probabilities calculated in Step 3 and sum over all \(n\) from 1 to 20. #

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