A. 3
B. 16
C. 75
D. 24
E. 26
A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box , if at least one black ball is to be included in the draw?
A. 32
B. 48
C. 64
D. 96
E. 55
A. 32
B. 48
C. 64
D. 96
E. 55
Discuss the answer using comment option. Official answer with explanation will be posted after 24 hrs
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Solution to 1st question
ReplyDeleteAns is E
3 way venn diagram intersection
total integers neither div by 2,3,5 = total nos - union of div by 2,3,5
= 101 - union of 3 way venn diag
Now,
Numbers divisible by 2 + Numbers Divisible by 3 + Numbers divisible by 5
-Numbers divisible by 2 and 3(ie 6)
-Numbers divisible by 3 and 5 (ie 15)
-Numbers divisible by 2 and 5(ie 10)
+Numbers divisible by 2,3 and 5(ie 30)
Now Numbers divisible by 2 = 51(Since it includes both 200 and 300)
Numbers divisible by 3 = 34 (ie,201-300)
Numbers divisible by 5 = 21 (inclusive of both 200 and 300)
Numbers divisble by 6 = 17
Numbers divisible by 10 = 11 (inclusive of both 200 and 300)
Numbers divisible by 15 = 7
Numbers divisble by 30 = 4
Therefore, union is
51+34+21-(17+11+7)+4
=110-35
=75
Hence, neither div by 2,3 or 5 is 101 - 75 =26
Solution to 2nd question
ReplyDeleteOA is C
Black Balls - 3
Non Black Balls - 2 +4 = 6
Combinations for at least 1 black from a total of 3 balls
1-black + 2-Other = 3C1 * 6C2 = 3 * 15 = 45
2-black + 1-Other = 3C2 * 6C1 = 3 * 6 = 18
3-black + 0-Other = 3C3 * 6C0 = 1 * 1 = 1
Total = 64.
Hi, Solution to 2nd question:
ReplyDeleteTotal ways = 9C3 = 84.
Ways of selecting no black ball = 6C3 = 20.
Finally, total ways of selecting atleast 1 Black ball = 84 - 20 = 64.