Question

A book has 350 pages. Some printing errors were observed on various pages and they were noted
down as follows:
No. of errors 0 1 2 3 4 5 6
No. of pages 75 65 55 60 33 40 22
Find the probability that a page selected at random will have
i. No errors
ii. More than 3 errors
iii. Only 3 errors

Answer 1

Answer:

i. No. of pages = 350 ( no. of possible outcomes)

No. of pages with no errors = 75 ( no. of favourable outcomes)

Therefore, P.E ( getting page with no errors ) = 75/350

→ 3/14 or 0.214

ii. No. of pages = 350

No. of pages with more than 3 errors = 94 (33+40+22)

Therefore, P.E ( getting page with more than 3 errors) = 95/350

→ 19/70 or 0.272

iii. No. of pages = 350

No. of pages with only 3 errors = 60

Therefore, P.E (getting page with only 3 errors) = 60/350

→ 6 / 35 or 0.171

Step-by-step explanation:

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