# Thread: help with probability!!

1. ## help with probability!!

I need help with the following questions, i attempted to solve them, and i also put my answers here too.
When you are answering them, please show all your steps as i am kinda lost in this field.

1. suppose S = {1,2,3} and P({1,2}) = 1/3 and P({2,3}) = 2/3
compute P({1}), P({2}), P({3})
P({1}) + P({2}) + P({3}) = 1 <--- 1 being 100% and 0 being 0%
P({1}) + P({2}) = 1/3
P({2}) + P({3}) = 2/3
a) P({1}) + (1/3 - P({1})) + P({3}) = 1
P({3}) = 2/3
b) P({1}) + (2/3 - P({3})) + P({3}) = 1
P({1}) = 1/3
so then 1/3 + P({2}) + 2/3 = 1
so P({2}) = 0
therefore: P({1}) = 1/3, P({2}) = 0, P({2}) = 2/3

2. Suppose A1 watches the six o'clock news a 2/3 of the time and watches eleven o'clock news 1/2 of the time and watches both news 1/3 of the time. For a randomely selected day what is the probability that A1 watches only the six o'clock new?s, and whats the probability that A1 watches neither news?
chance for 6oclock = 2/3
chance for 11oclock news = 1/2 => complementary is 1/2
for only six o'clock news
has to watch the 6oclock news which is 2/3, and not watch the 11oclock news which is 1/2
2/3*1/2 = 1/3
so only watching 6 o'clock news is 1/3 chance...
for watching neither
avoid watching 6o'clock news which is 1/3 (complementary) and avoid watching eleven o'clock news 1/2 (complementary)
so its
1/3*1/2 = 1/6
so watching neither is 1/6 chance...

3. Suppose your right knee is sore 15% of the time and your left knee is sore 10% of the time. What is the largest possible percentage that at least 1 knee is sore? What is the smallest possible percentage that at least 1 knee is sore?
Largest
15%+10% = 25%
Smallest
0.15*0.10 = 0.015
so 1.5%

4. Suppose a card is randomely chosen from a standard 52 card deck.
What is the probability that the card is a jack or a club (or both) ?
chance of jack = 4/52
chance of club = 13/52
chances of club or jack = 4/52 + 13/52 = 17/52
chances of jack and club is 1/52

5. Suppose 55% of students are female and 45% are male.
44% of females have long hair and 15% of males have long hair.
What is the probability that a random student will either be female or have long hair (or both)?
chance of female: 55%
chance of long hair: (55%*44%) + (45%*15%) = 0.2420 + 0.0675 = 0.3095 = 30.95%
chance of female or long hair = 55% + 30.95% = 85.95%
chance of female with long hair: 0.2420 = 24.2%

2. Hello, Sneaky!

4. Suppose a card is randomely chosen from a standard 52 card deck.
What is the probability that the card is a Jack or a Club (or both) ?

You started correctly, though . . .

There are 4 Jacks: $J\heartsuit,\,J\spadesuit,\, J\diamondsuit,\,\boxed{J\clubsuit}$

. . $P(J) \:=\:\dfrac{4}{52}$

There are 13 Clubs: $A\clubsuit,\,2\clubsuit,\,3\clubsuit,\,4\clubsuit, \,5\clubsuit,\,6\clubsuit,\,7\clubsuit,\,8\clubsui t,\,9\clubsuit,\,10\clubsuit,\,\boxed{J\clubsuit}, \,Q\clubsuit,\,K\clubsuit$

. . $P(\clubsuit) \:=\:\dfrac{13}{52}$

$\text{But }\,P(J\text{ or }\clubsuit) \;=\;\dfrac{16}{52}$

There are only 16 cards that will "win the bet".

You counted the $J\clubsuit$ twice.

3. ok i understand why its 16/52.
This also means that the other 4 questions were correct?
But now i have some other concerns.
This is a question from my book.

a guy has 3 cards
1 red on both sides
1 black on both sides
1 red / black

he puts 1 card on the table, and u only see the top side which is red.
what are the chances of the other side being red
my guess is 50/50, but the book says 2/3 using conditional probability.
Why is it 2/3??

Also a question about number 5
Like u said in number 4, i counted the jack of clubs twice, so in question 5 when adding 55% and 30.95% to get chance of female or a long hair person, doesn't part of the long hair people already count in the chance of females?