# Math Help - need help with conditional probability

1. ## need help with conditional probability

hi all

Two digits are chosen at randomfrom a table of random numbers containing digits 0,1,2....9. Find the probability that
(a) the sum of the two digits is greater than 9 given that the first number is 3
(b) the second number is 2 given that the sum of the two munber is greater than 7
(c) the first number is 4 given that the difference between the two numbers is 4.

this is whati said:
let A denote " the sum of two digits chosen is greater than nine"
let B denote "the first digit chosen is 3"

P(B) = 9/10

but i dont know the possibility space for A
i know that P(A given B) = P(A n B)/P(B)

so can u please give me a push in the rite direction. and a general short explaination please.

Thank you so much

2. Hello, helloworld101!

The problem has a rather small sample space.
It is easier to simply LIST the outcomes.

Two digits are chosen at random from a table of random numbers
. . containing digits: 0, 1, 2. ... 9.

Find the probability that

(a) the sum of the two digits is greater than 9, given that the first number is 3
The first number is a 3.
Sample space: . $\bigg\{(3,0), (3,1), (3,2) (3,3), (3,4), (3,5), (3,6), (3,7), (3,8), (3,9)\bigg\}$ . . . 10 outcomes

The sum is greater than 9: . $\bigg\{(3,7), (3,8), (3,0)\bigg\}$ . . . 3 outcomes.

Therefore: . $P(\text{sum}> 9\:|\:\text{1st is 3}) \;=\;\frac{3}{10}$

(b) the second number is 2, given that the sum of the two munber is greater than 7
The sum is greater than 8.

Sample space: . $\begin{Bmatrix}(0,8), (0,9) \\ (1,7), (1,8), (1,9) \\
(2,6), (2,7), (2,8), (2,9) \\ \vdots \\ (9,0), (9,1), (9,2), (9,3), \hdots (9,9) \end{Bmatrix}$
. . . 64 outcomes

The second number is 2: . $\bigg\{(6,2), (7,2), (8,2), (9,2)\bigg\}$ . . . 4 outcomes.

Therefore: . $P(\text{2nd is 2}\:|\:\text{sum} > 7) \;=\;\frac{4}{64} \;=\;\frac{1}{16}$

(c) the first number is 4, given that the difference between the two numbers is 4.
Difference of the numbers is 4.

Sample space: . $\begin{Bmatrix}(0,4), (1,5), (2,6), (3,7), (4,8), (5,0) \\ (4,0), (5,1), (6,2), (7,3), (8,4), (9,5) \end{Bmatrix}$ . . . 12 outcomes

First number is 4: . $\bigg\{(4,8), (4,0)\bigg\}$ . . . 2 outcomes

Therefore: . $P(\text{1st is 4}\:|\:\text{diff is 4}) \;=\;\frac{2}{12} \;=\;\frac{1}{6}$