Question

Two fair dice, each with faces numbered 1, 2, 3, 4, 5 and 6, are rolled together and the sum of the numbers on the faces is observed. This process is repeated till the sum is either a prime number or a perfect square. Suppose the sum turns out to be a perfect square before it turns out to be a prime number. If p is the probability that this perfect square is an odd number, then the value of 14p is

Answer 1

Two fair dice, each with faces numbered 1, 2, 3, 4 , 5 and 6 are rolled together and the sum of the numbers on the faces is observed. this process is repeated till the sum is either a prime number or a perfect square.

To find : if p is the probability that this perfect square is an odd number, then the value of 14p.

solution : sum of numbers will be [2 , 12]

so, there are two perfect square ; 4 and 9.

then ordered pairs ; (1, 3), (3, 1), (2,2) , (6, 3), (3, 6), (4, 5) and (5, 4)

so p(perfect square), p₁ = 7/36

p(perfect square is odd), p₂ = 4/36 [ n(E) = (6,3), (3,6) , (5,4) and (4,5) ]

now prime numbers ; 2 , 3, 5 , 7 or 11

so order pairs ; (1, 1), (1, 2), (2, 1), (1,4), (4, 1), (2, 3), (3, 2) , (1, 6), (6, 1), (2, 5), (5, 2) , (4, 3), (3, 4) , (5, 6) and (6, 5)

P(prime) = 15/36

now P(prime or square) = P(square) + P(prime)

= 7/36 + 15/36 = 22/36

so, P'(prime or square) , p'= 1 - 22/36 = 14/36

now required probability = [p₁ + p'p₁ + p'²p₁ + p'³p₁ + .....]/[p₂ + p'p + p'²p₂ + p'³p₂ + ......]

= (p₁/p₂)

= (4/36)/(7/36)

= 4/7

Therefore 14p = 14 × 4/7 = 8

Answer 2

The sum of the numbers on the faces ranges from $1+1=2$ to $6+6=12$ both inclusively.

The only perfect squares between 2 and 12 are 4 and 9.

Possible pairs of numbers on the dice whose sum is 4,

$\longrightarrow E(4)=\{(1,\ 3),\ (2,\ 2),\ (3,\ 1)\}$

No. of possible pairs,

$\longrightarrow n(E(4))=3$

Possible pairs of numbers on the dice whose sum is 9,

$\longrightarrow E(9)=\{(3,\ 6),\ (4,\ 5),\ (5,\ 4),\ (6,\ 3)\}$

No. of possible pairs,

$\longrightarrow n(E(9))=4$

We see $E(4)$ and $E(9)$ are disjoint.

Then, no. of possible pairs of numbers on the dice whose sum is a perfect square (either 4 or 9),

$\longrightarrow n(E(4\ or\ 9))=n(E(4)\cup E(9))$

$\longrightarrow n(E(4\ or\ 9))=n(E(4))+n(E(9))-n(E(4)\cap E(9))$

$\longrightarrow n(E(4\ or\ 9))=3+4-0$

$\longrightarrow n(E(4\ or\ 9))=7$

The set $E(4\ or\ 9)$ is the sample space as it contains total possible outcomes (probability of the sum being perfect square).

• $n(S)=7$

The set $E(9)$ is the favourable event as it contains favourable outcomes (probability of the sum being odd perfect square).

• $n(E)=4$

Then, probability of the sum being odd perfect square is,

$\longrightarrow p=\dfrac{n(E)}{n(S)}$

$\longrightarrow p=\dfrac{4}{7}$

Therefore,

$\longrightarrow\underline{\underline{14p=8}}$

Hence 8 is the answer.