Question bank/JEE Main/Mathematics
JEE MainMathematicsToughProbability
A person throws dice, one the common cube and the other regular
terahedron, the number on the lowest face being taken in the case of a
tetrahedron. The chance that the sum of numbers thrown is not less than
5 is.,d O;fDr nks iklksa dks QSadrk gS ftlesa,d lk/kkj.k ?ku rFkk nwljk
Show answer and solution
Total number of combinations of numbers on the cube and the
tetrahedon = 6 × 4 = 24
Favourable number of ways of getting a sum not less than 5
= sum of coefficients of x^6, x^7,.........x^10 in the product
= (x + x^2 + x^3 + x^4 + x^5 + x^6) (x + x^2 + x^3 + x^4)
= (x^2 + 2x^3 + 3x^4 + 4x^5 + 4x^6 + 4x^7 + 3x^8 + 2x^9 +
x^10) = 4 + 4 + 4 + 3 + 2 + 1 = 18
Required
probability.
gy VsVªkgsMksu ij dqy = 6
× 4 = 24
5 ls de ugha gksus vuqdwy
mRikn esa x^6, x^7,.........x^10 ds xq.kkad
= (x + x^2 + x^3 + x^4 + x^5 + x^6) (x + x^2 + x^3 + x^4)
= (x^2 + 2x^3 + 3x^4 + 4x^5 + 4x^6 + 4x^7 + 3x^8 + 2x^9 +
x^10) = 4 + 4 + 4 + 3 + 2 + 1 = 18
laHkkouk .
tetrahedon = 6 × 4 = 24
Favourable number of ways of getting a sum not less than 5
= sum of coefficients of x^6, x^7,.........x^10 in the product
= (x + x^2 + x^3 + x^4 + x^5 + x^6) (x + x^2 + x^3 + x^4)
= (x^2 + 2x^3 + 3x^4 + 4x^5 + 4x^6 + 4x^7 + 3x^8 + 2x^9 +
x^10) = 4 + 4 + 4 + 3 + 2 + 1 = 18
Required
probability.
gy VsVªkgsMksu ij dqy = 6
× 4 = 24
5 ls de ugha gksus vuqdwy
mRikn esa x^6, x^7,.........x^10 ds xq.kkad
= (x + x^2 + x^3 + x^4 + x^5 + x^6) (x + x^2 + x^3 + x^4)
= (x^2 + 2x^3 + 3x^4 + 4x^5 + 4x^6 + 4x^7 + 3x^8 + 2x^9 +
x^10) = 4 + 4 + 4 + 3 + 2 + 1 = 18
laHkkouk .
Source: Grass-G / M-21 Mathematics Paper 19 Dec Hindi Eng.docx
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