a box contains card number of 1, 3, 5, 7 – – – to 215 if one card is drawn at random find the probability of getting – (a) perfect square number {b} a perfect cube number (c) a number divisible by 7 (d) three digit number (e) a number divisible by 6
Answers
Answered by
1
Answer:
(a) 8/215 (b)4/215 (c)22/215 (d)2/215 (e) 25/215
Step-by-step explanation:
(a)no of favourable outcomes = 8
total number of outcomes =215
therefore, P(getting a perfect square number)=8/215
(b)no of favourable outcomes = 4
total number of outcomes =215
total number of outcomes =215P(getting a perfect cube number) = 4/215
(c) no of favourable outcomes = 22
total number of outcomes =215
therefore,P(getting anumber divisible by 7) = 22/215
(d)no of favourable outcomes =2
total number of outcomes =215
therefore,P(getting a three digit number) = 2/225
(e)no of favourable outcomes = 25
total number of outcomes =215
therefore,P(getting a divisible by 6) = 25/215
I hope this would help you my friend
Similar questions