First slide

Introduction to probability

Question

Two numbers b and c are chosen at random (with replacement from the numbers 1,2,3,4,5,6,7,8 and 9). The probability that $x^{2} + bx + c > 0$ for all $x \in R$ is

Moderate

A

17/123
B

32/81
C

82/125
D

45/143

Solution

Here, $x^{2} + bx + c > 0 \forall x \in R$
$\begin{array}{ll} ∴ D < 0 \\ \Rightarrow b^{2} < 4 c \end{array}$
value of b possible values of c
$\begin{array}{l} 1 1 < 4 c \Rightarrow c > \frac{1}{4} \Rightarrow {1, 2, 3, 4, 5, 6, 7, 8, 9} \\ 2 4 < 4 c \Rightarrow c > 1 \Rightarrow {2, 3, 4, 5, 6, 7, 8, 9} \\ 3 9 < 4 c \Rightarrow c > \frac{9}{4} \Rightarrow {3, 4, 5, 6, 7, 8, 9} \\ 4 16 < 4 c \Rightarrow c > 4 \Rightarrow {5, 6, 7, 8, 9} \\ 5 25 < 4 c \Rightarrow c > 625 \Rightarrow {7, 8, 9} \\ 6 36 < 4 c \Rightarrow c > 9 Impossible \end{array}$
7 impossible
8 impossible
9 impossible
Number of favorable ca8e8 =9 + 8 + 7+6+3=32
Total ways=9x9=81
Required probability = 32/81

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