Question

Question :

What is the probability of sum 10 with 5 dice?​


#Exam - Time.


#No - Spams!!

Answer 1

$\huge{\boxed{\tt{Answer:-}}}$

If five dice are thrown, the event space contains 6⁵ event points.

The event 'sum 10' contains points of the type (x₁, x₂, x₃, x₄, x₅) where x₁, x₂, x₃, x₄, x₅ can take values from the set {1, 2, ... , 6} and Σ x₁ = 10.

$\huge{\boxed{\tt{Explanation:-}}}$

Now, number of outcomes of this type,

=coefficient of x¹⁰ in the multinomial expansion of (x¹ + x² ... + x⁶)⁵

Now, (x¹ + x² + ... + x⁶)⁵

= x⁵ (1 + x + ... + x⁵)⁵

= x⁵ {(1 - x⁶)/(1 - x)}⁵

= x⁵ (1 - x⁶)⁵ (1 - x)⁻⁵

= x⁵ (1 - ⁵C₁ x⁶ + ...)

{1 + 5x + ... + (5.6.7.8.9)/5! x⁵ + ...}

∴ coefficient of x¹⁰ in the above expansion

= (5.6.7.8.9)/5!

= 15120/120

$\large{\boxed{\tt{= 126}}}$

∴ probability of 'sum 10' with five dice

= 126/6⁵

$\large{\boxed{\tt{= 7/432}}}$

Answer 2

The event 'sum 10' contains points of the type (x₁, x₂, x₃, x₄, x₅) where x₁, x₂, x₃, x₄, x₅ can take values from the set {1, 2, ... , 6} and Σ x₁ = 10.

Now, number of outcomes of this type,

=coefficient of x¹⁰ in the multinomial expansion of (x¹ + x² ... + x⁶)⁵

Now, (x¹ + x² + ... + x⁶)⁵

= x⁵ (1 + x + ... + x⁵)⁵

= x⁵ {(1 - x⁶)/(1 - x)}⁵

= x⁵ (1 - x⁶)⁵ (1 - x)⁻⁵

= x⁵ (1 - ⁵C₁ x⁶ + ...)

{1 + 5x + ... + (5.6.7.8.9)/5! x⁵ + ...}

∴ coefficient of x¹⁰ in the above expansion

= (5.6.7.8.9)/5!

= 15120/120

∴ probability of 'sum 10' with five dice

= 126/6⁵