Thread: Coin tossing and Cards Experiment Which I Find Hard To Solve

Hello,

This may seem little ridiculous but I have problem with solving these two problems. Will be really good if someone can help me to solve this. As I am sure I will get a same model questions in my exams next week. any help will be great.

Question 1:

Ten unbiased coins are tossed simultaneously. Find the probability of getting atleast seven heads.


Question 2:

Four Cards are drawn at Random from the pack of 52 cards. Find the probability that.

a. These are a king, a queen, a jack and an ace

b. Two are Kings and two are queens

c. Two are black and two are red

d. There are two cards of hearts and two cards of diamond.

Any help to solve these couple of questions will be highly appreciated.

Thanks

Originally Posted by chakru

Hello,

This may seem little ridiculous but I have problem with solving these two problems. Will be really good if someone can help me to solve this. As I am sure I will get a same model questions in my exams next week. any help will be great.

Question 1:

Ten unbiased coins are tossed simultaneously. Find the probability of getting atleast seven heads.


Question 2:

Four Cards are drawn at Random from the pack of 52 cards. Find the probability that.

a. These are a king, a queen, a jack and an ace

b. Two are Kings and two are queens

c. Two are black and two are red

d. There are two cards of hearts and two cards of diamond.

Any help to solve these couple of questions will be highly appreciated.

Thanks

Can we see some attempt plz?

Ok for the first question.. I did like this

P(7) = p(X=7) = (1/2)^7
P(8) = p(X=8) = (1/2)^8
P(9) = p(X=9) = (1/2)^9
P(10) = p(X=10) = (1/2)^10

Question 2:

a. These are a king, a queen, a jack and an ace

4/52 * 4/51 * 4/50 * 4/49 =

b. Two are Kings and two are queens

4/52 * 3/51 * 4/50 * 3/49 =

c. Two are black and two are red

26/52 * 25/51 * 26/50 * 25/49 =

d. There are two cards of hearts and two cards of diamond.

13/52 * 12/51 * 13/50 * 12/49 =

I am not sure that's the right answer though..

I think your Question 2 is correct.

I believe you should do question 1 by calculating the number of combinations of 7 in 10 (120), plus 8 in 10 (45), plus 9 in 10 (10), plus 10 in 10 (1), over the total number of combinations possible (1024)