1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Probability of equal roots in equation

  1. Sep 18, 2016 #1
    1. The problem statement, all variables and given/known data
    A quadratic eqn of form ax2 + bx + c = 0 is selected. The values of a, b and c are distinct and selected from 1, 2, 3, 4, 6, 8, 9. What is probability of chosen equation to have equal roots?

    2. Relevant equations
    root = (-b +/- Sqrt(b2-4ac)) / (2a)
    For equal roots b2 = 4ac

    3. The attempt at a solution

    well i found out that for b2 = 4ac, the conditions are
    b=6, a = 1, c = 9.
    b=6, a = 9, c = 1.

    Probability = no. of likely events/total no. of events.
    Probability of A and B to occur = P(A)*P(B)

    Case 1---
    So probability of value of a to be 6 is 1/7. (as total 7 numbers are there)
    Probability of value of b to be 1 is 1/7
    Probability of c to be 9 is 1/7.

    Case 2---
    Likewise Probability of a to be 6 is again 1/7
    P(b=9) = 1/7
    P(c=1) = 1/7

    P(equal roots) = P(case 1) + P(case 2)
    = 1/73 + 1/73

    But in solutions it goes like: there are two ways in which condition can be achieved for equal roots.
    Total ways = 7P3.
    So P(equal roots) = 2/(7P3)

    Why is my method wrong? And why should it be 7P3 and not 7C3?
  2. jcsd
  3. Sep 18, 2016 #2


    User Avatar
    Gold Member

    Since a must be 6, there are not 7 possibilities for b and c. And after b is chosen, there are not 6 possibilities for c.

    Edit: P instead of C because it matters which number is chosen for which letter. If we are picking a,b,c then 6,1,9 is not equivalent to 9,1,6.
  4. Sep 18, 2016 #3
    Thanks for the reply. So it would be 1/7 * 1/6 * 1/5 + (same as first term)

    But when we say two cards were pulled from a pack of 52, we say 52C2, we don't go as 52C1 * 51C1.
  5. Sep 18, 2016 #4


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You've solved the problem where ##a, b, c## need not be distinct, as you've included all the cases where ##a=b## etc.
  6. Sep 18, 2016 #5


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    We'll that depends. Most card games don't make a distinction based on the order you get the cards. So: 6-Hearts + 8-Clubs is the same as 8-Clubs + 6-Hearts.

    The number of 2-card hands is, therefore, ##(52 \times 51)/2 = \binom{52}{2}##

    But, if you have a situation where the order the cards get dealt matters, then 6-Hearts + 8-Clubs is different from 8-Clubs + 6-Hearts and the number of 2-card draws is: ##52 \times 51 = 52P2##

    In your problem ##a =1, b = 2, c = 3## is very different from ##a = 3, b = 2, c =1##.
  7. Sep 18, 2016 #6


    User Avatar
    Gold Member

    Well we might. But having got 52C1 x 51C1 we would have calculated 52P2 and would then divide by 2P1 (or 2!) to get 52C2 if that was what we wanted.

    Depending on what you mean by "two cards were pulled from a pack" you may want 52C2 or 52P2. As pointed out before, in the quadratic problem, you wanted P rather than C, so 52C1 x 51C1 was right, giving you 52P2.
  8. Sep 18, 2016 #7


    User Avatar
    Gold Member

    This seems to say that either it is impossible to select the same number twice (equivalent to choice without replacement), or that any duplicate choices are ignored, which boils down to the same thing, since the only results you look at are the ones that you would have got without replacement.

    You therefore seem to be interpreting the question as, "a,b and c are selected from {1,2,3,4,6,8,9}: what is the probability that they are distinct and give equal roots?"
    I'm not saying this is an impossible interpretation, but it seems to me an odd way of stating the question if that was the intended interpretation.
  9. Sep 18, 2016 #8

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    I deleted my previous post, because I had missed the word "distinct" in the OP, and so my post was misleading and irrelevant.
  10. Sep 19, 2016 #9
    Thanks a lot for the replies. I guess the distinct matters like 6,1,9 isn't same as 9,1,6 but when you're picking cards it doesn't matter that you get 9, Ace, 6 or 6, Ace, 9 since in cards its the combination that's more important than order.
  11. Sep 20, 2016 #10


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I assume you mean order matters, not distinct matters.
    If you decide to pick a number and assign it to a, pick another and assign it to b, and a third to c (say) then the order matters.
    If you just pick the three numbers first then decide how to assign them, the order in which you pick them does not matter, but it does matter how you assign them. Either way, it's 7P3.
  12. Sep 20, 2016 #11
    Yeah...thanks for correcting me...
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted