# Thread: Coins tosses and probability students pass an exam

1. ## Coins tosses and probability students pass an exam

I need help for two tasks:

1.
Twogroups, A and B wrote an exam. There are 60 students in group A and 70 studentsin group B. The probability of A students’ pass is 0.3 and the probability of Bstudents' pass is 0.5. What is the expected number of students who will take the exam? What is the probability that only one studentto pass the exam?

2. Aplayer tosses two coins, betting 50 cent. He wins 20 cent for each fallen head.What is the expected winnings in a single toss? What should be the amount of winningsin order the game to be fair (not possible tendency for the player to win orlose)?

2. ## Re: Coins tosses and probability students pass an exam

Is this a translation from another language?

3. ## Re: Coins tosses and probability students pass an exam

Yes. Which part is not clear?

4. ## Re: Coins tosses and probability students pass an exam

Originally Posted by dtzoneva
I need help for two tasks:

1. Two groups, A and B wrote an exam. There are 60 students in group A and 70 studentsin group B. The probability of A students’ pass is 0.3 and the probability of Bstudents' pass is 0.5. [/FONT]What is the expected number of students who will take the exam[FONT="Calibri"]?
??? All 130 students take the exam! If you mean the expected number of students who pass the exam, the expected number of students in group A who pass is (0.3)60= 18 and the expected number of students in group B who pass is (0.5)70= 35. The total number of students that are expected to pass is 18+ 35= 53.

What is the probability that only one studentto pass the exam?
In order that only one student pass the exam either one student in A passes and all other 59 students in group A fail and all 70 students in group B fail or one student in group B passes while all other 69 students in group B fail and all 60 students in group A fail. The probability of the first is $\displaystyle (0.3)(0.7^{59})(0.5^{70})$ and the probability of the second is $\displaystyle (0.7^{60})(0.5)(0.5^69)= (0.7^{60})(0.5^{70})$. The probability that, in both groups only one student will pass is the sum of those two.

2) A player tosses two coins, betting 50 cent. He wins 20 cent for each fallen head.What is the expected winnings in a single toss? What should be the amount of winningsin order the game to be fair (not possible tendency for the player to win or lose)?
I presume you mean "a single toss" of both coins. For each coin the probability of heads or tails is 1/4. With two coins, the probability of two heads is 1/4, the probability of one head is 1/2, and the probability of no heads is 1/4. The expected winnings is 40(1/4)+ 20(1/2)+ 0(1/4)= 10+ 10= 20 cents. Since you pay 50 cents to play, you would expect to lose 30 cents each time. Suppose you were to get "a" cents each head rather than 20 cents. Then your expected winnings would be 2a(1/4)+ a(1/2)+ 0(1/4)= a. In order to "break even" that must be the 50 cents you paid so a= 50 cents.