# Thread: Two students are randomly picked from a class of 10 boys and 5 girls..

1. ## Two students are randomly picked from a class of 10 boys and 5 girls..

Two students are randomly picked from a class of 10 boys and 5 girls. What is the probability that the second student picked is a girl, given that the two has at least 1 girl?

the answer is 1/6 but i dont know how to get that..

Explanation would be appreciated thanks!

2. I must be doing something wrong since I got 7/12

3. Originally Posted by qkqh
Two students are randomly picked from a class of 10 boys and 5 girls. What is the probability that the second student picked is a girl, given that the two has at least 1 girl?

the answer is 1/6 but i dont know how to get that..

Explanation would be appreciated thanks!
I get 7/12 too..

$\displaystyle P(B|A)=\dfrac{P(B\cap A)}{P(A)}$

A = there's at least one girl chosen
B = the second one is a girl

$\displaystyle P(A) = 1 - P(\overline{A}) = 1 - \dfrac{\binom{10}{2}}{\binom{15}{2}} = \dfrac{4}{7}$

$\displaystyle P(B\cap A) = P(B) = \dfrac{10}{15} \cdot \dfrac{5}{14} + \dfrac{5}{15} \cdot \dfrac{4}{14} = \dfrac{1}{3}$

Note: You could also get P(B) = 1/3 by reasoning that the probability that the second one is a girl is the same as the probability that the first one is a girl.

4. I used order (BG means boy first, then girl)

$\displaystyle P(BG \cup GG| BG\cup GG\cup GG)$

$\displaystyle ={P(BG)+P(GG)\over P(BG)+P(GG)+P(GG)}$

$\displaystyle ={ ({10\over 15}) ({5\over 14}) +({5\over 15}) ({4\over 14}) \over ({10\over 15}) ({5\over 14}) +({5\over 15}) ({10\over 14}) + ({5\over 15}) ({4\over 14})}$

multiply everywhere by (15)(14) leads to 70/120 or 7/12.

5. Hello, qkqh!

I solved it with my usual baby-talk approach.

And I too have no idead how they got $\displaystyle \frac{1}{6}$

Two students are randomly picked from a class of 10 boys and 5 girls.
What is the probability that the second student picked is a girl,
given that the two has at least one girl?

We choose 2 students from a class of 15.
. . There are: .$\displaystyle _{15}C_2 \:=\:{15\choose2} \:=\:210$ outcomes.

$\displaystyle \begin{array}{ccccc} (1) &\text{Girl-Girl} & 5\cdot4 &=& 20 \\ (2) & \text{Girl-Boy} & 5\cdot10 &=& 50 \\ (3) & \text{Boy-Girl} & 10\cdot5 &=& 50 \\ (4) & \text{Boy-Boy} & 10\cdot 9 &=& 90 \\[-3mm] & & & &-- \\[-2mm] & & & & 210 \end{array}$

Since the two students include at least one girl,
. . the sample space is: .$\displaystyle (1) + (2) + (3) \:=\:20 + 50 + 50 \:=\:120$

The second student is a girl: .$\displaystyle (1) + (3) \:=\:20 + 50 \:=\:70$

Therefore: .$\displaystyle P(\text{2nd is Girl}\,|\,\text{at least one girl}) \;=\;\dfrac{70}{120} \;=\;\boxed{\frac{7}{12}}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

"They" could get $\displaystyle \frac{1}{6}$ with: .$\displaystyle \frac{35}{210}$ or $\displaystyle \frac{20}{120}$

. . neither of which makes any sense.