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CBSE Class 12-science Answered

A bag contains 7 white , 5 Black and 4 red balls. 4 balls are drawn without replacement. Find the probability that atleast 3 balls are black.
Asked by sachinkumarmishra1 | 24 Jan, 2020, 11:10: AM
answered-by-expert Expert Answer
A bag contains 7 white , 5 Black and 4 red balls. 4 balls are drawn without replacement. To find the probability that atleast 3 balls are black.
 
T o t a l space n o. space o f space b a l l s space equals space 7 plus 5 plus 4 equals 16
P left parenthesis a t space l e a s t space 3 space b a l l s space a r e space b l a c k right parenthesis space equals space P left parenthesis e x a c t l y space 3 space b a l l s space a r e space b l a c k right parenthesis thin space plus space P left parenthesis A l l space 4 space b a l l s space a r e space b l a c k right parenthesis

A s space t h e r e space a r e space 11 space n o n minus b l a c k space b a l l s space a n d space t h e s e space b a l l s space c a n space b e space a r r a n g e d space i n space 4 space w a y s space a m o n g space t h e m s e l v e s
space P left parenthesis e x a c t l y space 3 space b a l l s space a r e space b l a c k right parenthesis space equals space fraction numerator C presubscript 1 presuperscript 5 over denominator C presubscript 1 presuperscript 16 end fraction cross times fraction numerator C presubscript 1 presuperscript 4 over denominator C presubscript 1 presuperscript 15 end fraction cross times fraction numerator C presubscript 1 presuperscript 3 over denominator C presubscript 1 presuperscript 14 end fraction cross times fraction numerator C presubscript 1 presuperscript 11 over denominator C presubscript 1 presuperscript 13 end fraction cross times 4 space
space P left parenthesis A l l space 4 space b a l l s space a r e space b l a c k right parenthesis space equals space fraction numerator C presubscript 1 presuperscript 5 over denominator C presubscript 1 presuperscript 16 end fraction cross times fraction numerator C presubscript 1 presuperscript 4 over denominator C presubscript 1 presuperscript 15 end fraction cross times fraction numerator C presubscript 1 presuperscript 3 over denominator C presubscript 1 presuperscript 14 end fraction cross times fraction numerator C presubscript 1 presuperscript 2 over denominator C presubscript 1 presuperscript 13 end fraction
therefore space P left parenthesis a t space l e a s t space 3 space b a l l s space a r e space b l a c k right parenthesis space equals space open parentheses 5 over 16 cross times 4 over 15 cross times 3 over 14 cross times 11 over 13 cross times 4 close parentheses plus open parentheses 5 over 16 cross times 4 over 15 cross times 3 over 14 cross times 2 over 13 close parentheses
space space space space space equals space 5 over 16 cross times 4 over 15 cross times 3 over 14 cross times 1 over 13 open parentheses 11 cross times 4 plus 2 close parentheses equals 5 over 16 cross times 4 over 15 cross times 3 over 14 cross times 1 over 13 open parentheses 46 close parentheses equals 23 over 364
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