Question

At a card party, 2 door prizes are awarded by drawing 2 names randomly from a box of 40 names, including Joe and Pat. What's the probability that joe is not chosen for either prize?

Answer 1

Is this 39/40 x 38/39, 95%?

Answer 2

ANd what is the probability Pat wins 1 prize?

Answer 3

Is the 2nd one 1/40+1/39?

Answer 4

@Nnesha

Answer 5

@Loser66

Answer 6

So there are 40 cards, only 2 are picked. 2 of them are two people. So Joe is one of them, what is the fraction?

Answer 7

@dtan5457

Answer 8

joe is not chosen for either prize.

Answer 9

@Sup_Peeps

Answer 10

for a moment i thought 1-p would do

Answer 11

for which part?

Answer 12

prob that joe is not chosen

Answer 13

but they choose twice

Answer 14

@sleepyjess

Answer 15

@dtan5457 Your solution to the first part (p = 0.95) is correct. The second part can be solved by using combinations:
\[\large P(Pat\ wins)=\frac{39C1}{40C2}\]

Answer 16

They both are extremely close in terms of the answer, why combinations, though?

Answer 17

The question deals with sampling without replacement, when a sample of size n is taken from a population containing a items of one particular type, and b of others. The probability that the sample contains exactly x of the particular type is given by the relevant term in the hypergeometric distribution.
\[\large p(x)=\frac{\left(\begin{matrix}a \\ x\end{matrix}\right)\left(\begin{matrix}b \\ n-x\end{matrix}\right)}{\left(\begin{matrix}a+b \\ n\end{matrix}\right)}\]

Answer 18

Using the formula for the first part we get:
\[\large p(0)=\frac{1C0 \times 39C2}{40C2}=0.95\]