If $D_{1} a n d D_{2}$ be two biased dice such that the probabilities of getting six on a single throw are $\frac{4}{6}$ and $\frac{2}{6}$ respectively. Suppose $α$ is a number of sixes that appear when $D_{1}$ is a thrown twice, independently and suppose $β$ is a number of sixes that appear when $D_{2}$ is thrown twice. Then probability that the roots of the quadratic polynomial $x^{2} - α x + β$ are real and equal is

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{20}{81}$

$\frac{40}{81}$

$\frac{1}{2}$

$\frac{1}{4}$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

(2) $α^{2} = 4 β C a s e I : α = 0, β = 0 P (x = r) = \sum_{r = 0}^{n} n_{c_{r}} p^{r} q^{n - r}$
$= (\frac{1}{3} \times \frac{1}{3}) \times (\frac{2}{3} \times \frac{2}{3}) = \frac{4}{81}$ Case II:
$P (α = 2, β = 1) = (\frac{2}{3} \times \frac{2}{3}) \times (\frac{1}{2} \times \frac{2}{3} + \frac{2}{3} \times \frac{1}{3}) = \frac{16}{81}$
Required probability = $\frac{20}{81}$