sis

Hassin Atahifh

04/15/2018

IM311

Question 1 (12pts): A box contains 7 blue, 8 red and 6 green balls. You pick one ball randomly. What is the probability that the ball that you picked is neither green nor red?

In this case, the question is the probability of picking a blue ball. That is;

Probability (Neither red or green) = Probability (Blue)

There are 7 blue balls, out 21 balls (7+8+6 = 21), hence P (Blue) = 7/21 = 1/3

Question 2 (15pts): A box contains 4 blue, 5 red and 6 white balls. We decide to take 3 balls randomly so what is the probability that all of them are red?

Selecting 3 balls from 15 balls (4 + 5 + 6 = 15), we have 15C3 = 455 ways disregarding the order in which they occur. If all balls are red, then we have 4C1*3C1*2C1 = 288 ways of doing so. That gives the probability of drawing all three red balls as 288/455 = 0.6330

Question 3 (15pts): Assume that a coin is tossed and a die is rolled at the same time, calculate the probability that the coin shows a head and the die shows an even number?

P (Head) = ½

P (2 or 4 or 6) = 1/6 + 1/6 + 1/6 = 3/6 = ½

Hence P (H or Even Number) = ½ * ½ = ¼

Question 4 (14pts): A box contains 4 apples and 4 carrots. You eat 3 of them (by randomly choosing). What is the probability that you eat 2 apples and 1 carrot?

Probability of picking two apples = P (Apples) = 4/8+3/7 = 13/14

Probability of picking one carrot = P (Carrot) = 4/8 = 1/2

P (Two apples and one carrot) = 13/14*1/2 = 13/28

Question 5(14pts): You toss 3 unbiased coins (fair coin). Please calculate the probability of getting at most 2 heads

To better view this solution, we better draw a sample space first which is as follows

(HHH, HHT, HTH, THH, TTT, TTH, THT, HTT).

You can have a total of 8 outcomes from which we can count the ones with at most two heads.

P (At most two heads) = favorable outcomes/total outcomes = 7/8. Note that even the outcomes which have no head areincluded because at most means getting zero, one or two heads.

Question 6 (20pts): 5 persons are selected from a group of 6 women and 7 men, to form a committee, where at least 3 men are selected in the committee. Please calculate in how many different ways it can be done?

The total ways of forming a committee are given by 13C5 = 1,287. After this, we can calculate total ways of forming a committee with 3 men, 2 men, 1 man and zero men, with women topping the committee to be five members. This is given by;

7C3*6C2+7C4*6C1+7C5*6C0 = 35*15 + 35*6 + 21*1 = 756 ways of picking at least three men in a committee of five members.

Question 7 (10pts): In how many ways, the letters of the word 'SUPPORT' can be arranged? (The order is not important)

The letters are 7. The table below shows the number of each distinct letters

S	1
U	1
P	2
O	1
R	1
T	1

Then this gives us;

7!/1!1!2!1!1!1!1! = 2,520 ways of rearranging the word SUPPORT.