Thread: problems related to probability ?

1. problems related to probability ?

Hi All,

I am stuck in the below problem, so need assistance.

Q1 - A person draws a card from a pack of 52 , replaces it and shuffles it. He continues doing it until he draws a heart.what is the probability he has to make 3 trails ?

Book Answer is 9/64

Q2 - If 4 whole numbers was taken at random and multiplied together, what is the chance that the last digit in the product is 1,3,7 or 9 ?

Any help would be greatly appreciated.

Thanks,
Ashish

2. 9/64 is correct
this is a geometric random variable
P((no heart on first selection)(no heart on second selection)(heart on third selection))
=P(no heart on first selection)P(no heart on second selection)P(heart on third selection)
by independence
=(3/4)(3/4)(1/4)

3. I don't really understand (2)
Is there a range of numbers?
Can we pick 1 through 100,000,000?

4. Originally Posted by a69356
Hi All,

I am stuck in the below problem, so need assistance.

Q1 - A person draws a card from a pack of 52 , replaces it and shuffles it. He continues doing it until he draws a heart.what is the probability he has to make 3 trails ?

Book Answer is 9/64

Q2 - If 4 whole numbers was taken at random and multiplied together, what is the chance that the last digit in the product is 1,3,7 or 9 ?

Any help would be greatly appreciated.

Thanks,
Ashish
Q2. Matheagle is quite right, you have to make some assumption about the probability space from which the numbers are drawn. We will just assume that the space is large enough that the probability that the remainder when a number is divided by 10 is evenly distributed in the range 0, 1, 2, ..., 9. I.e., if X is one of our random numbers, the probability that $X \equiv n \pmod{10}$ is 1/10 for n = 0, 1, 2, ..., 9.

Suppose the 4 numbers are $X_1, X_2, X_3, \text{ and } X_4$. Their product ends in 1, 3, 7, or 9 if it does not end in 0, 2, 4, 5, 6, or 8; in other words, if it is not divisible by 2 or 5. Since 2 and 5 are primes, the product is not divisible by 2 or 5 if none of $X_1, X_2, X_3, \text{ and } X_4$ is divisible by 2 or 5. For each i,

$P(X_i \text{ is not divisible by 2 or 5}) = P(X_i \equiv 1, 3, 7, \text{ or } 9 \pmod{10}) = 4/10$.

The $X_i$'s are independent, so the probability that $X_1 X_2 X_3 X_4$ ends in 1, 3, 7, or 9 is

$P(X_1, X_2, X_3 \text{ and } X_4 \text{ are not divisible by 2 or 5}) = (4/10)^4$.