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(D) The total number of past working days is $200$.Let $E$ be the event that tomorrow's output will have at least one defective part.From the table,th

Vedclass · Accepted Answer

$0.75$

Vedclass · Answer

(D) The total number of past working days is $200$.Let $E$ be the event that tomorrow's output will have at least one defective part.From the table,the number of days with $0$ defective parts is $50$.The number of days with at least one defective part is the total number of days minus the number of days with $0$ defective parts.Number of days with at least one defective part $= 200 - 50 = 150$.The probability $P(E)$ is given by the ratio of the number of days with at least one defective part to the total number of days.$P(E) = \frac{	ext{Number of days with at least one defective part}}{	ext{Total number of days}}$$P(E) = \frac{150}{200} = \frac{3}{4} = 0.75$.

Number of defective parts	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$	$10$	$11$	$12$	$13$
Days	$50$	$32$	$22$	$18$	$12$	$12$	$10$	$10$	$10$	$8$	$6$	$6$	$2$	$2$

Blood group	$A, AB, B, O$
Number of students	$10, 13, 12, 5$

Number of defective bulbs	$0$	$1$	$2$	$3$	$4$	$5$	$6$	more than $6$
Frequency	$400$	$180$	$48$	$41$	$18$	$8$	$3$	$2$

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