Explanation
Step 1: Identify the Sample Space
The first three prime numbers are 2, 3, and 5.
Since the coefficients a,b, and c must be distinct (a=b=c), they represent a permutation of the set {2,3,5}.
Total number of ways to arrange these 3 numbers for a,b, and c is:
Total Outcomes (n(S))=3!=3×2×1=6
Let's list all 6 possible combinations of (a,b,c):
(2,3,5)
(2,5,3)
(3,2,5)
(3,5,2)
(5,2,3)
(5,3,2)
Step 2: Condition for Real Roots
For a quadratic equation to have real roots, its discriminant (D) must be greater than or equal to zero:
D=b2−4ac≥0
⟹b2≥4ac
Step 3: Check Favorable Outcomes
Let's test the condition b2≥4ac for each of the 6 possible arrangements:
Case | (a,b,c) | b2 | 4ac | Is b2≥4ac? |
1 | (2,3,5) | 32=9 | 4×2×5=40 | 9≥40 ❌ (False) |
2 | (2,5,3) | 52=25 | 4×2×3=24 | 25≥24 (True) |
3 | (3,2,5) | 22=4 | 4×3×5=60 | 4≥60 ❌ (False) |
4 | (3,5,2) | 52=25 | 4×3×2=24 | 25≥24 (True) |
5 | (5,2,3) | 22=4 | 4×5×3=60 | 4≥60 ❌ (False) |
6 | (5,3,2) | 32=9 | 4×5×2=40 | 9≥40 ❌ (False) |
Number of favorable outcomes (n(E))=2 (which are (2,5,3) and (3,5,2)).
Step 4: Calculate the Probability
The probability P(E) that the roots are real is given by:
P(E)=n(S)n(E)=62=31
Correct Answer:
(b) 31
