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Hint: Probability of event E = $\dfrac{n(E)}{n(S)}=\dfrac{\text{Favourable cases}}{\text{Total number of cases}}$ where S is called the sample space of the random experiment. Find n (E) and n (S) and use the above formula to find the Probability.

Complete step-by-step answer:

Let E be the event: The watch chosen is defective.

Since there are 50 defective watches the total number of cases favourable to E = 50.

Hence, we have n(E) = 50.

The total number of ways in which we can choose the wrist watches = 500.

Hence, we have n(S) = 500.

Hence, P (E) = $\dfrac{50}{500}=0.1$

Hence the Probability that the chosen wristwatch is defective = 0.1

Note:

[1] It is important to note that drawing uniformly at random is important for the application of the above problem. If the draw is not random, then there is a bias factor in drawing, and the above formula is not applicable. In those cases, we use the conditional probability of an event.

[2] The Probability of an event always lies between 0 and 1

[3] The sum of Probabilities of an event E and its complement Eâ€™ is 1.

i.e. $P(E)+P(E')=1$

Hence, we have $P(E')=1-P(E)$. This formula is applied when it is easier to calculate P(Eâ€™) instead of P(E).

Complete step-by-step answer:

Let E be the event: The watch chosen is defective.

Since there are 50 defective watches the total number of cases favourable to E = 50.

Hence, we have n(E) = 50.

The total number of ways in which we can choose the wrist watches = 500.

Hence, we have n(S) = 500.

Hence, P (E) = $\dfrac{50}{500}=0.1$

Hence the Probability that the chosen wristwatch is defective = 0.1

Note:

[1] It is important to note that drawing uniformly at random is important for the application of the above problem. If the draw is not random, then there is a bias factor in drawing, and the above formula is not applicable. In those cases, we use the conditional probability of an event.

[2] The Probability of an event always lies between 0 and 1

[3] The sum of Probabilities of an event E and its complement Eâ€™ is 1.

i.e. $P(E)+P(E')=1$

Hence, we have $P(E')=1-P(E)$. This formula is applied when it is easier to calculate P(Eâ€™) instead of P(E).

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