Ps PROBABILITIES 4Gmat 1

Question

HELLO

here it is : IF 2 numbers are chosen at random from the set '{1,2,3....10}with replacement what is the probability that the roots of the quadraticequation x^2+ax+b=0 has real roots ? plz explain yours works..... thanks

chets · Answer

9/10

(oops! looks like 9/10 is off by miles!)

vikramm · Answer

yaieks again.... Mandy, where are you getting these problems??? This one took me almost 5 mins just to think how to approach it... 

Here we go... I think I have the right answer. 

Roots of the Equation are: ( (-a+Sqrt(a^2-4b))/2a, (-a-Sqrt(a^2-4b))/2a)
So for the roots to be real... anything under that sqrt sign cannot be negative. 

is. a^2 - 4b >= 0
or a^2 >= 4b

Given set of numbers to pick (I guess this list is to pick values for a and b) is 1 through 10... so lets turn these values in terms of a^2 and 4b to get the number of favorable outcomes
Set = {1,2,3,4,5,6,7,8,9,10}
a^2 = {1,4,9,16,25,36,49,64,81,100}
4b = {4,8,12,16,20,24,32,36,40} 

From the above for a^2 = {7,8,9,10), all 10 from the list of 4b's can be picked = 4*10 = 40 
Similarly, 
a^2 = 4 pick 4b = 4, or 1*1 = 1
a^2 = 9, pick 4,8 or 1*2 = 2
a^2 = 16, pick 4,8,12,16 or 4
a^2 = 25, pick 4,8,12,16,20,24 or 6
a^2 = 36, pick 4,8,12,16,20,24,32,26 or 8
Also, a^2 = 1, b cannot be picked. 

Total Outcomes = All 10 for a and All 10 for b = 10*10 = 100
Favorable outcomes = 40+1+2+4+6+8 = 62

Probability = 62/100 or 0.62 

Is that correct?

duttsit · Answer

x^2+ax+b=0 will have imaginary root when:
a 2 < 4b (ax^2 + b*x + c will have imaginary root if b^2 - 4 * a * c < 0)

lets now inumerate what all combinations of a & b satisfies this relationship:

i) a = 1, b = 1,2,3...,10
ii) a = 2, b = 2,3,...,10
iii) a = 3, b = 3,4,...,10
iv) a=4,b = 5,6,..,10
v) a=5, b = 7,8,9,10
vi) a = 6, b = 10

Finding probability of all of these and adding gives probability of solution sets when there is imaginary root for the equation x ^ 2 + a x + b = 0

p (imaginary root) = 1/10 * 10 /10 + 1/10 * 9/10 + 1/10 * 8/10 + 1/10 * 6/10 + 1/10 * 4/10 + 1/10 * 1/10

pr p = 38 /100 or 19/50

therefore, probability for real root = 
p(real root) = 1 - 19 / 50 = 31/50 or 0.62

ranga41 · Answer

I get 0.62 too

from formula for roots, to get real roots a^2 >= 4b

if a = 1 any b will give imaginary root
if a = 2 (1 alone will give real root)
if a = 3 (1,2 will give real root)
if a = 4 (then 1,2,3,4)
if a = 5 (1 thru 6)
if a = 6 (t thru 9)
for all other a (1 thru 10 will give real root)

so 62 combination out of 100 gives real root...

mandy · Answer

Vikramm yes you got the right answer those problems comes from a software about permutations and probabilities made by 4gmat I recently purchased : https://www.4gmat.com/aboutus.shtml in case you want to have a look I always take the most difficult and challenging ones because I think it is the way to get true insight on a topic and on test day, we will find ets problems easy regards Mandy

Ps PROBABILITIES 4Gmat 1

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards