A bag contains 4 red balls, 6 black balls and 8 yellow balls. 3 balls are picked at random. Find the probability that at least one of them is yellow.
A - 25/34
B - 27/34
C - 29/34
D - 21/34
Answers
Answered by
0
Answer:
21/34
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Answered by
1
Answer:
No. of red balls =4
No. of yellow balls =5
No. of green balls =6
Total no. of balls=15
Reqd. Combination is G,G,(R/Y)
So ⁶C2×⁹C1
=>6!/2!4! ×9!/8!1!
=>15×9
=>135
There are 135 ways in which the desired combination can be obtained
Total no. Of ways is
¹⁵C3
=15!/3!12!
=13×14×15/3
=910
.’. PROBABILITY =135/910
=27/182
Your answer should be 27/182
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