Let and denote the outcome of three independent rolls of a fair tetrahedral die, whose four faces are marked 1, 2, 3, 4. If the probability that has all real roots is then is equal to

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
Let $a, b$ and $c$ denote the outcome of three independent rolls of a fair tetrahedral die, whose four faces are marked 1, 2, 3, 4. If the probability that $a x^{2} + b x + c = 0$ has all real roots is $\frac{m}{n}, g c d (m, n) = 1,$ then $m + n$ is equal to _______ . [2024]

Ans.
(19)

We have, $a x^{2} + b x + c = 0$

For real roots, $b^{2} - 4 a c \geq 0$ ...(i)

$a, b, c \in {1, 2, 3, 4}$ ...(ii)

Ordered triplet $(a, b, c)$ satisfying (i) and (ii) are

$(1, 2, 1), (1, 3, 1), (2, 3, 1), (1, 3, 2), (1, 4, 1), (1, 4, 2), (2, 4, 1), (2, 4, 2), (4, 4, 1), (1, 4, 4), (3, 4, 1), (1, 4, 3)$

i.e. total 12 favourable outcomes.

Total number of outcomes = $4 \times 4 \times 4 = 64$

$∴$ $Required probability = \frac{12}{64} = \frac{3}{16} = \frac{m}{n}$ $(∵ Given)$

Here, $m + n = 3 + 16 = 19$

Want Us to Email you About Special Offers & Updates?