Question

There are 3 black and 7 white balls in the box, identical in touch. You draw 3 balls randomly. If you find 1 black ball among the drawn ones, you will win $1, if you find 2 black balls, you win $2 and you will win $10, if all 3 are black. What are the chances to win at least 1 buck?

Answer 1

P(drawing a black ball) = 3/10 = 0.3
Since you draw 3 balls randomly, let n = 3
and we have the binomial distribution X~B(3,0.3)
P(X≥1) = 1 - P(X=0) = 0.657

Answer 2

\[P(X \ge1) = 1 - P(X=0)\]

Answer 3

If \(X\) is the random variable describing your winnings, you can have \(X=1, 2, 10\). You want \(P(X \ge 1)\) which mean \(P(X=1)+P(X=2)+P(X=10)\)
\[ \large P(X=1)=\frac{{3\choose 1}{7 \choose 2}}{{10 \choose 3}} \] And try the same reasoning for the other 2. Essentially for X=1, I considered that you choose 1 black ball from 3, and then 2 white balls from 7.
I think :o

Answer 4

\[\large P(1\ black)=\frac{C(3, 1)\times C(7, 2)}{C(10, 3)}=0.525\]
\[\large P(2\ black)=\frac{C(3, 2)\times C(7, 1)}{C(10, 3)}=0.175\]
\[\large P(3\ black)=\frac{C(3,3)\times C(7, 0)}{C(10, 3)}=0.00833\]
The probability of winning at least $1 is the sum of the above 3 values of probability.

Answer 5

Note: The events 'draw one black', 'draw two black' and 'draw three black' are mutually exclusive. Therefore the probability of drawing 1 or 2 or 3 black is the sum of the individual probabilities.

Answer 6

The simplest way to find the solution is to find the probability of drawing zero black. Then subtract that probability from 1.
\[\large P(0\ black)=\frac{C(3, 0)\times C(7, 3)}{C(10, 3)}=0.29166\]
\[\large P(1\ or\ 2\ or\ 3\ black)=1-0.29166=0.70833\]