Question

Three dice are thrown. The probability of getting a sum which is a perfact square, is -

• A. $\frac { 2 } { 5 }$
• B. $\frac { 9 } { 20 }$
• C. $\frac { 1 } { 4 }$
• D. None of these

Answer 1

Answer: (D)

None of these

Answer 2

n(S) = 6 × 6 × 6. Clearly, the sum varies from 3 to 18, and among these 4, 9, 16 are perfect squares. The number of ways to get the sum 4 = the number of integral solutions of x₁ + x₂ + x₃ = 4

Where 1 £ x₁ £ 6, 1 £ x₂ £ 6, 1 £ x₃ £ 6

= coefficient of x⁴ in (x + x² +…. + x⁶)³

= coefficient of x in $\left( \frac { 1 - x ^ { 6 } } { 1 - x } \right) ^ { 3 }$= coefficient of x in (1 – x⁶)³ . (1 – x)^–3 = ³C₁ Similarly, the number of ways to get the sum 9

= –3 × 1 + ⁸C₆ = 28 – 3 = 25.

The number of ways to get the sum 16

= coefficient of x¹³ in (1 – x⁶)³ . (1 – x)^–3

= coefficient of x¹³ in (1 – x⁶)³ . (1 – x)^–3 = ³C₁ = coefficient of x¹³ in (1 – 3x⁶ + 3x¹² –x¹⁸) (²C₀ + ³C₁x + ⁴C₂x² + ….)

= ¹⁵C₁₃ – 3 × ⁹C₇ + 3 × ³C₁ = 105 – 108 + 9 = 6 \ n (5) = 3 + 25 + 6 = 34.

So, P (5) = $\frac { 34 } { 6 \times 6 \times 6 }$ = $\frac { 17 } { 108 }$ . Hence (4) is correct answer