A sample of 4 different calculators is randomly selected from a group containing 44 that are defective and 22 that have no defects. What is the probability that all four of the calculators selected are defective?

To find the probability that all four of the calculators selected are defective, we can use the concept of combinations and probabilities. Given: - Total number of defective calculators = 44 - Total number of calculators with no defects = 22 - Total number of calculators = 44 + 22 = 66 - Number of calculators selected = 4 The probability of selecting a defective calculator in the first draw is 44/66. After selecting the first defective calculator, there are 43 defective calculators left out of 65 total calculators. The probability of selecting a defective calculator in the second draw is 43/65. Similarly, the probabilities for the third and fourth draws are 42/64 and 41/63, respectively. The probability of selecting all four defective calculators is the product of these probabilities. Let's calculate this probability. Calculate the value by following steps: - step0: Calculate: \(\left(\frac{44}{66}\right)\times \frac{43}{65}\left(\frac{42}{64}\right)\times \frac{41}{63}\) - step1: Reduce the fraction: \(\frac{2}{3}\times \frac{43}{65}\left(\frac{42}{64}\right)\times \frac{41}{63}\) - step2: Reduce the fraction: \(\frac{2}{3}\times \frac{43}{65}\times \frac{21}{32}\times \frac{41}{63}\) - step3: Multiply the terms: \(\frac{86}{195}\times \frac{21}{32}\times \frac{41}{63}\) - step4: Multiply the terms: \(\frac{301}{1040}\times \frac{41}{63}\) - step5: Reduce the numbers: \(\frac{43}{1040}\times \frac{41}{9}\) - step6: Multiply the fractions: \(\frac{43\times 41}{1040\times 9}\) - step7: Multiply: \(\frac{1763}{9360}\) The probability that all four of the calculators selected are defective is approximately 0.1883.