i need help with a statistics probability word problem

Question

Four balls are picked from a box containing 7 white, 6 black, 5 red and 4 green balls. Find the probability that

(a) all balls are of different colors.

(b) all balls are of the same color.

Answer 1

Just some quick notes to get you on track (I hope)...

Are the balls being picked and discarded? Or picked and replaced? The former changes the odds of the next ball (one less of the color picked, one less total balls), while the latter the odds are unchanged (same number of colors, same number of total balls). I'd guess you're replacing them, which makes the problem a bit easier.

There is a probability for picking a color, based on number of balls of that color, and total balls in the box.

P(W) = 7/22

P(B) = 6/22 = 3/11

P(R) =

P(G) =

If P(W1) = 7/22

then with replacement, probability is unchanged so P(W2) = 7/22, P(B2) = 6/22

but without replacement, probability is changed so P(W2) = 6/21, P(B2) = 6/21

With these four probabilities you can answer questions like: what is the probability of picking a red ball? P(R). Or what is the probability of picking a white ball AND a green ball? P(W) x P(G). Or what is the probability of picking a black ball OR a red ball? P(B) + P(R). Or even what is the probability of picking a white ball four times? P(W1) x P(W2) x P(W3) x (PW4)

But your questions are a little harder. :-(

(a) How can you can get all different colors?

Pick White first. Then Black. Then Red. Then Green.

Or White, black, green, red.

Or white, red, black, green

etc

WBRG

WBGR

WRBG

WRGB

WGBR

WGRB

That's 6 ways, starting with white. I think I included them all. Yes?

P(WBRG) = P(W) x P(B) x P(R) x P(G) = ?

this is probability of picking one combination of first white ball choices

P(WBRG) = P(WBGR) = P(WRBG) etc so all combos have same probability

and P(all white first combinations) = P(combo1) OR P(combo2) OR P(combo3) etc ie 6 x P(WBRG)

But you also have (very similar to above)...

starting with black BWRG etc

and

starting with red RWBG etc

and

starting with green GWBR etc

Is P(GWBR) same as P(WBRG)?

(b) How could you get all same colors?

Here's white: P(W)⁴ = (7/22)⁴

Or, if no replacement, 7/22 x 6/21 x 5/20 x 4/19 (multiply because it's White AND white AND white AND white)

BUT... you could pick 4 x W, or 4 x B, or 4 x R, or 4 x G. Here you ADD the all color probabilities because it's all white OR all black OR all red OR all green.

So now you can calculate P(C1=C2=C3=C4)

Answer 2

22 balls

Probability Draw white first

7/22 - leaves 21 balls, 6 white

Probability Not draw white second

15/21 - leaves 20 balls, 6 white

Probability Not draw white third

14/20 - leaves 19 balls, 6 white

Probability Not draw white fourth

13/19

probability draw white first and no other balls match: 7/22 * 15/21 * 14/20 * 13/19

Repeat computations for other three colored balls then add the probabilities together

b) probability draw all whites

7/22 * 6/21 * 5/20 * 4/19

repeat computations for other three colored balls then add the probability together

Answer 3

a) Since we only have 4 colors of balls and want 4 different colors, we will have one of each color.

Thus, 7 * 6 * 5 * 4/C(7+6+5+4,4) = 7*6*5*4/C(22,4) = 840/7315 = 24/209

b) all 4 are white, black, red, or green

(C(7,4)+C(6,4)+C(5,4)+C(4,4))/C(22,4) =(35+15+5+1)/7315 = 56/7315 = 8/1045