Here is the exact wording in the problem given.
Numbers from 1 to 10 are printed on a fair 10-faced dice. The dice is thrown 4 times.
Find the probability the result has 3 consecutive prime numbers or no prime number shows up 3 times consecutively.
Here's my approach so far. I am a Math Olympiad enthusiast so you can use any such math.
In a throw,
P(prime) = 4/10 = 2/5
P(not prime) 6/10 = 3/5
Next
P(3 consecutive primes) = 2 x (2/5)^3 x (3/5) = 48/625
because 2 configurations ppp_ or _ppp so we need to multiply by 2.
also
P(no primes appear 3 times consecutively)
= 1 – P (some prime number appears 3 times consecutively)
= 1 – 2 x (4/10) (1/10)^2 x (9/10)
= 1241/1250
Because 2 configurations as before (so multiply by 2)
(4/10) is probability of first prime followed by (1/10)^2 as probability of next two consecutive but "same" prime with the first while (9/10) is probability for selection of remaining throw to be any number except the repeated prime here.
I wondered if there is an overlap to the cases above by Principle of Inclusion Exclusion which we need to subtract after adding the numbers above
P(Overlap) —> add the following breakdown of cases:
P(all three primes different)
= 2 x (4/10)(3/10)(2/10) x 10/10 – (4/10)(1/10)^3
2 configurations PPP_ or PPP
(4/10) represents probability choice for first prime
(3/10) represents probability choice for second prime which must be different from the first
(2/10) represents probability choice for third prime which must be different from both earlier ones.
(10/10) represents that the remaining number "" can be either a prime or a non-prime.
We minus (4/10)(1/10)^3 because that part is repeated in both configurations above as case with PPPP.
+
P(two primes only exactly the same which we count in two configuration sets as counting is not symmetrical within both)
= P(configurations P1P1P2_ or P1P2P1_ or _P2P1P1 or _P1P2P1)
- P(Configurations P2P1P1_ or _P1P1P2)
= 4 (4/10)(1/10)(3/10)(8/10) + 2 x (4/10)(1/10)(3/10)(9/10)
In calculator
P(overlap) = 243/2500
So Probability we seek is
= 48/625 + 1241/1250 – 243/2500
= 2431/2500
Problem: The answer key (no solution) given states that the answer is 253/625. How to get it please and where did I go wrong? Any help is appreciated, thank you
Best Answer
I must say that the question has been very poorly posed.
By doing reverse engineering, I have concluded that what they are actually asking is the probability that in $4$ throws, you get at least $3$ primes or $3$ non-primes (i.e. by "no prime" they mean non-prime)
Leaving out the denominator (which will always be $625$)
number of favorable cases $$= 2(2^3*3) +2^4 + 2(3^3*2) +3^4 = 253$$ $\quad\quad\quad and\; Pr= \dfrac{253}{625} $