Probability

Probability Level 3

There are 5 balls, with numbers 3, 3, 4, 4, and 4 written on each. You randomly pick one of the balls. If 3 is written on it, you throw a die 3 times. If 4 is written on it, you throw a die 4 times. The probability that the sum of the numbers from the die is 10, is p q \frac{p}{q} , for coprime positive integers p and q . What is p + q ?


The answer is 587.

Tyler I am lying 2 by Tyler I am lying 2 , 88, South Korea

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2 solutions

Pop Wong
Feb 14, 2021

Let

  • X i X_i be the random variable of the outcome of rollling a die at i i trail
  • S n S_n be the sum of ∑ i = 1 n X i \sum_{i=1}^{n} X_i
  • e.g. S 3 S_3 be the sum of X 1 + X 2 + X 3 X_1 + X_2 + X_3

Note:

  • Rolling a standard die 2 2 times, the sum will be from 2 2 to 12 12 and the distribution is symmetric.
  • P r ( S 2 = 2 ) = 1 36 = P r ( S 2 = 12 ) Pr(S_2=2) = \cfrac{1}{36} = Pr(S_2=12)
  • P r ( S 2 = 3 ) = 2 36 = P r ( S 2 = 11 ) Pr(S_2=3) = \cfrac{2}{36} = Pr(S_2=11)
  • P r ( S 2 = 10 ) = 3 36 ( ( 4 , 6 ) , ( 5 , 5 ) , ( 6 , 4 ) ) Pr(S_2=10) = \cfrac{3}{36} \hspace{10mm}( (4,6), (5,5), (6,4) )

P r ( 4 ≤ S 2 ≤ 9 ) = 1 − P r ( S 2 = 2 ) − P r ( S 2 = 3 ) − P r ( S 2 = 10 ) − P r ( S 2 = 11 ) − P r ( S 2 = 12 ) = 36 − 1 − 2 − 3 − 2 − 1 36 = 27 36 Pr( 4 \leq S_2 \leq 9) =1 - Pr(S_2=2)- Pr(S_2=3) - Pr(S_2=10) - Pr(S_2=11)- Pr(S_2=12) = \cfrac{36 - 1 -2 -3 -2 -1 }{36} = \cfrac{27}{36}

P r ( S 3 = 10 ) = ∑ k = 4 9 P r ( S 2 = k ) × P r ( X 3 = 10 − k ) = P r ( 4 ≤ S 2 ≤ 9 ) × 1 6 = 27 36 × 1 6 = 27 216 Pr(S_3 = 10 ) = \sum_{k=4}^{9} Pr(S_2=k)\times Pr(X_3= 10-k) = Pr( 4 \leq S_2 \leq 9) \times \cfrac{1}{6} = \cfrac{27}{36} \times \cfrac{1}{6} = \boxed{ \cfrac{27}{216} }

Note:

  • Rolling a standard die 3 3 times, the sum will be from 3 3 to 18 18 and the distribution is symmetric.
  • P r ( S 3 = 10 ) = P r ( S 3 = 11 ) Pr(S_3 = 10 ) = Pr(S_3 = 11 )
  • P r ( 4 ≤ S 3 ≤ 9 ) = P r ( 12 ≤ S 3 ≤ 17 ) Pr( 4 \leq S_3\leq 9 ) = Pr( 12 \leq S_3 \leq 17 )

P r ( 4 ≤ S 3 = X 1 + X 2 + X 3 ≤ 9 ) = 1 − P r ( S 3 = 3 ) − P r ( S 3 = 18 ) − P r ( S 3 = 10 ) − P r ( S 3 = 11 ) 2 Pr( 4 \leq S_3=X_1 + X_2 + X_3 \leq 9 ) = \cfrac{1 - Pr( S_3= 3 ) - Pr( S_3 = 18 ) - Pr(S_3 = 10 ) - Pr(S_3 = 11 ) }{2}

Cases P r ( S 3 = 3 ) Pr( S_3= 3 ) P r ( S 3 = 18 ) Pr( S_3 = 18 ) P r ( S 3 = 10 ) Pr(S_3 = 10 ) P r ( S 3 = 11 ) Pr(S_3 = 11 ) Total
P r Pr 1 216 \cfrac{1}{216} 1 216 \cfrac{1}{216} 27 216 \cfrac{27}{216} 27 216 \cfrac{27}{216} 56 216 \cfrac{56}{216}

P r ( 4 ≤ S 3 ≤ 9 ) = 80 216 Pr( 4 \leq S_3 \leq 9 ) = \cfrac{80}{216}

∴ P r ( S 4 = 10 ) = 80 216 × 1 6 \therefore \boxed{ Pr( S_4=10 ) = \cfrac{80}{216}\times \cfrac{1}{6} }

The probability of getting a sum equals 10 10

= 2 5 × P r ( S 3 = 10 ) + 3 5 × P r ( S 4 = 10 ) = 2 5 × 27 216 + 3 5 × 80 216 × 1 6 = 54 + 40 216 × 5 = 47 540 = \cfrac{2}{5} \times Pr(S_3 = 10 ) + \cfrac{3}{5} \times Pr(S_4=10)\\ = \cfrac{2}{5} \times \cfrac{27}{216} + \cfrac{3}{5} \times \cfrac{80}{216}\times \cfrac{1}{6}\\ = \cfrac{54+40}{216\times5}\\ = \cfrac{47}{540}

∴ p + q = 47 + 540 = 587 \therefore p+q = 47+540 = \boxed{587}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

i) If the number written on the ball is 3 (probability: 2/5)

possible numbers from the die are: (not considering the order)

(6, 3, 1), (6, 2, 2), (5, 4, 1), (5, 3, 2), (4, 4, 2), (4, 3, 3)

Therefore, the probability is

2/5 * (3! + 3 + 3! + 3! + 3 + 3) * (1/6)^3 = 1/20

ii) If the number written on the ball is 4 (probability: 3/5)

possible numbers from the die are: (not considering the order)

(6, 2, 1, 1), (5, 3, 1, 1), (5, 2, 2, 1), (4, 4, 1, 1), (4, 3, 2, 1), (4, 2, 2, 2), (3, 3, 3, 1), (3, 3, 2, 2)

Therefore, the probability is

3/5 * (4P2 + 4P2 + 4P2 + 4C2 + 4! + 4 + 4 + 4C2) * (1/6)^4 = 1/27

So, the answer is

1/20 + 1/27 = 47/540

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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