Math, asked by StrongGirl, 7 months ago

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

Answers

Answered by abhi178
20

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

Answered by shadowsabers03
12

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.

Similar questions