# sports probability

• Jan 17th 2010, 01:38 AM
thereddevils
sports probability
In a sports quiz , each participant is required to answer a question , 2 of which are on badminton and 3 on soccer . THe questions are kept in separate envelopes . These envelopes are numbered 1 to 5 . The four participants are A,B C and D . They are required to pick their own questions randomly .

For each participants , what is the probability that he will be the first to pick a question on badminton.

so this is what i have in mind , its either he picks badminton (B) first , then 2nd guy pick B , then the rest soccers (BBSS) , or (BSBS) or (BSSS)

there are 3 cases here , (2/5)(3/5)(3/5)(3/5)+(2/5)(3/5)(2/5)(3/5)+(2/5)(3/5)(3/5)(3/5) but the answer given is 3/10 . Erm , i cant see my mistake here .
• Jan 17th 2010, 06:39 AM
Soroban
Hello, thereddevils!

The problem is not stated clearly,
. . so my interpretation may be wrong.

Quote:

In a sports quiz, each participant is required to answer a question,
2 of which are on badminton and 3 on soccer.
The four participants are A, B, C and D.
They are required to pick their own questions randomly .

For each participant, what is the probability that he will be the first to pick a question on badminton?

I assume the following:

. . The participants pick one question from the five.

. . The selected question is not replaced.

. . The participants do their picking in random order.

With all events being equally likely, each participant has an equal chance
. . of being the first to choose a badminton question.

Then why isn't this true?
. . . $P(\text{first to pick badminton}) \;=\;\frac{1}{4}$

• Jan 17th 2010, 06:56 AM
P(1st person picks B)= $\frac{2}{5}=\frac{4}{10}$
P(B is picked 1st by the 2nd person)= $\frac{3}{5}\frac{2}{4}=\frac{3}{10}$
P(B is first picked by the 3rd person)= $\frac{3}{5}\frac{2}{4}\frac{2}{3}=\frac{2}{10}$
P(B is first picked by the 4th person)= $\frac{3}{5}\frac{2}{4}\frac{1}{3}\frac{2}{2}=\frac {1}{10}$