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CBSE Class 12-science Answered

Three dice are thrown simultaneously. Find the probability of getting a triplet of two's, if it is known that the sum of the numbers on the dice was six.   Thanks for your answers given on 8th March. However my Definite integral question was  integral subscript 0 superscript pi space fraction numerator space x space d x over denominator 1 space plus space e sin x end fraction space and the answer to this question is fraction numerator pi space cos to the power of negative 1 end exponent space x over denominator square root of 1 minus e squared end root end fraction However you have taken the integral as integral subscript 0 superscript pi fraction numerator x space d x over denominator 1 plus space e to the power of sin x end exponent end fraction     Request you to please provide the solution to the first stated definite integral. Thanks  
Asked by ahuja8087 | 09 Mar, 2017, 02:39: AM
answered-by-expert Expert Answer

Dear Student, esinx maybe a typo from where you have taken the question. Otherwise there is no such thing for CBSE XII Syllabus. I think it should be e^sinx. Kindly check. We have checked too. We have not come across esinx anytime. Or maybe it is, e^x sinx or something. But esinx we have not come across.

begin mathsize 16px style When space three space dice space are space thrown space simultaneously comma space total space number space of space elementary space events equals 6 cubed equals 216 Let space straight A space be space the space event space that space the space sum space of space the space numbers space on space the space dice space is space 6. rightwards double arrow straight A equals left curly bracket left parenthesis 1 comma space 2 comma space 3 right parenthesis comma space left parenthesis 1 comma space 3 comma space 2 right parenthesis comma space left parenthesis 2 comma space 3 comma space 1 right parenthesis comma space left parenthesis 2 comma space 1 comma space 3 right parenthesis comma space left parenthesis 3 comma space 1 comma space 2 right parenthesis comma space left parenthesis 3 comma space 2 comma space 1 right parenthesis comma space left parenthesis 1 comma space 1 comma space 4 right parenthesis comma space left parenthesis 1 comma space 4 comma space 1 right parenthesis comma space left parenthesis 4 comma space 1 comma space 1 right parenthesis comma space left parenthesis 2 comma space 2 comma space 2 right parenthesis right curly bracket rightwards double arrow straight P left parenthesis straight A right parenthesis equals 10 over 216 Let space straight B space be space the space event space of space getting space straight a space triplet space of space two apostrophe straight s. rightwards double arrow straight B equals left curly bracket left parenthesis 2 comma space 2 comma space 2 right parenthesis right curly bracket rightwards double arrow straight P left parenthesis straight B right parenthesis equals 1 over 216 And comma space straight P left parenthesis straight A intersection straight B right parenthesis equals 1 over 216 Hence comma space required space probability equals straight P left parenthesis straight B divided by straight A right parenthesis equals fraction numerator straight P left parenthesis straight A intersection straight B right parenthesis over denominator straight P left parenthesis straight A right parenthesis end fraction equals fraction numerator begin display style bevelled 1 over 216 end style over denominator begin display style bevelled 10 over 216 end style end fraction equals 1 over 10 end style

Answered by | 09 Mar, 2017, 08:12: PM

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