A jar contains....

Question

A jar contains 7 balls. Two of the balls are labeled 1, three are labeled 2 and two are labeled 3. Calculate the probability that the number on the second ball is greater than the number on the first ball.

My attempt taking into account combination:

(2C1)(3C1) drawing a 1 first and then drawing a 2. (2C1)(2C1) drawing a 1 first and then drawing a 3. (3C1)(2C1) drawing a 2 first and then drawing a 3.

(2)(3)+(2)(2)+(3)(2) = 16 ways to draw a greater number on the second draw.

Total ways of drawing balls (7C2) = 21

My answer: 16/21 = .761904762. But the book is giving me .380. Is my reasoning incorrect. Please help.

Answer 1

There are $7\cdot 6=42$ possible cases.

The number of favorable cases now.

• If we extract an $1$ first, there are $2\cdot(3+2)$ corresponding favorable cases.
• If we extract a $2$ first, there are $3\cdot 2$ corresponding favorable cases.
• No favorable cases if the $3$ comes first. Probability is thus

$$ \frac 1{42}\Big(\ 2(3+2)+3\cdot 2\ \Big) = \frac {16}{42} = \frac 8{21} = 0,\ 380952\ 380952\ 380952\ \dots\ . $$

Answer 2

$^7\mathrm C_2$ will count the ways to select two balls from seven.   You don't want just that.

You need to count the ways to select a first and a second ball from the seven.   Since order mattered to you in the numerator, so too it needs to matter here in the denominator.   That is $\def\cbinom#1#2{{^#1\mathrm C_#2}} \cbinom 72 2!$ or $\cbinom 71\cbinom 61$.

$$\dfrac{\cbinom 31\cbinom 21+\cbinom 21\cbinom 21+\cbinom 21\cbinom 31}{\cbinom 71\cbinom 61}=\dfrac{16}{42}$$

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Reality Check: By symmetry the probability that the second ball is less than the first must equal the probability the first ball is less than the second. So the probability for selecting equal numbers in the first and second draws should be $(42-2(16))/42$.   Using the above method we find:$$\dfrac{\cbinom 21\cbinom 11+\cbinom 31\cbinom 21+\cbinom 21\cbinom 11}{\cbinom 71\cbinom 61}=\dfrac{10}{42}$$