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Question Number 139697 by otchereabdullai@gmail.com last updated on 30/Apr/21

$$\mathrm{In}\mathrm{an}\mathrm{entrance}\mathrm{examination}\mathrm{each}$$$$\mathrm{question}\mathrm{has}\mathrm{four}\mathrm{possible}\mathrm{answers}.$$$$\mathrm{A}\mathrm{candidate}\mathrm{answers}\mathrm{three}\mathrm{question}$$$$\mathrm{by}\mathrm{guessing}.$$$$\mathrm{a}.\mathrm{find}\mathrm{the}\mathrm{probability}\mathrm{that}\mathrm{all}\mathrm{three}$$$$\mathrm{questions}\mathrm{are}\mathrm{wrong}.$$$$\mathrm{b}.\mathrm{find}\mathrm{the}\mathrm{probability}\mathrm{that}\mathrm{all}\mathrm{the}$$$$\mathrm{questions}\mathrm{are}\mathrm{correct}.$$

Answered by mr W last updated on 30/Apr/21

$$a.$$$$p=\frac{3\times 3\times 3}{4\times 4\times 4}=\frac{27}{64}$$$$b.$$$$p=\frac{1\times 1\times 1}{4\times 4\times 4}=\frac{1}{64}$$

Commented by otchereabdullai@gmail.com last updated on 30/Apr/21

$$\mathrm{Prof}\mathrm{W}\mathrm{you}\mathrm{are}\mathrm{supper}!$$

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