Answer: 0.69727
Step-by-step explanation:
Given: Sample size : n= 4
Number of defective calculators = 12
Number of non-defective calculators = 36
Total calculators = 12+36=48
Let X = Number of defective calculators.
The probability that a least one of the calculators is defective will be :
[tex]P(X\geq1)=P(X=1)+P(X=2)+P(X=3)+P(X=4)\\\\=\dfrac{^{12}C_1\times\ ^{36}C_{3}}{^{48}C_4}+\dfrac{^{12}C_2\times\ ^{36}C_{2}}{^{48}C_4}+\dfrac{^{12}C_3\times\ ^{36}C_{1}}{^{48}C_4}+\dfrac{^{12}C_4\times\ ^{36}C_{0}}{^{48}C_4}\\\\=\dfrac{12\times7140}{194580}+\dfrac{66\times\ 630}{194580}+\dfrac{220\times\ 36}{194580}+\dfrac{495\times\ 1}{194580}\approx0.69727[/tex]
Hence, the required probability = 0.69727
