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CBSE Class 12-science Answered

There are 2n+1 well suffled cards consecutively numbered. 3 cards are drawn at random from it find the probability that the three cards drawn are in arithematic progression.
Asked by mishrapragu1998 | 13 Mar, 2016, 12:13: AM
answered-by-expert Expert Answer
begin mathsize 12px style Total space number space of space ways space of space selecting space 3 space cards space out space of space 2 straight n plus 1 space cards space is space straight C presuperscript 2 straight n plus 1 end presuperscript subscript 3 If space we space select space the space 1 st space card space numbered space 1 comma space to space get space an space arithmetic space triplet comma space the space second space number space can space be space 2 comma space 3 comma.... space straight n minus 1 space left parenthesis till space the space middle space term right parenthesis So space the space second space number space can space be space chosen space in space straight n minus 2 space ways. Similarly space for space 2 space as space first space term space of space straight A. straight P. comma space we space have space straight n minus 3 space choices space for space 2 nd space card. Also space for space 3 space as space first space term space of space straight A. straight P. comma space we space have space straight n minus 3 space choices space for space second space card. So space the space total space number space of space ways equals open parentheses straight n minus 2 close parentheses plus 2 open parentheses straight n minus 3 close parentheses plus 2 open parentheses straight n minus 4 close parentheses plus.... plus 1 equals open parentheses straight n minus 2 close parentheses plus 2 open square brackets open parentheses straight n minus 3 close parentheses plus open parentheses straight n minus 4 close parentheses plus... plus 1 close square brackets equals open parentheses straight n minus 2 close parentheses plus 2 fraction numerator open parentheses straight n minus 3 close parentheses open parentheses straight n minus 2 close parentheses over denominator 2 end fraction equals open parentheses straight n minus 2 close parentheses plus open parentheses straight n minus 3 close parentheses open parentheses straight n minus 2 close parentheses equals open parentheses straight n minus 2 close parentheses squared So space probability equals fraction numerator open parentheses straight n minus 2 close parentheses squared over denominator straight C presuperscript 2 straight n plus 1 end presuperscript subscript 3 end fraction end style
Answered by satyajit samal | 14 Mar, 2016, 01:23: PM
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