Probability Help: Understanding Basic Concepts and Solving Practice Problems

[F.B]

I have lots of question on probability because my teacher makes us learn from the book so i need help.

My first question is:

1. Suppose the two joker cards are left in a standard deck of cards. One of the jokers is red and the other is black. A single card is drawn is from the deck of 54 cards,but not returned, and then a second card is drawn. Determine the probability of drawing the red joker or a red ace on either draw.

This is the answer my teacher told me but i have no idea how to get 51/54 or the 50/53 so can anyone please explain this to me.
1- (51/54)*(50/53)= 1- 425/477 = 52/477

2. A basketball player has a success rate of 80% for shooting free throws.
Calculate the following probabilities.
i) She will make 3 out of 5 attempts
ii)She will miss all 5
iii)She will make the first 3 and miss the next 2.

This one i have no idea how to do because we are still on condition probability so i don't kno how to do permutations or anything

My last question is this:

An airplane can make a safe landing if at least half of its engines are working. Suppose that engine failures are independant events. Determine whether a two-engine plane is safer that a four engine plane if the chance that an engine fails is 1 in 2.

i also have no idea how to do this questions either.

So please can anyone explain step by step on how to do these questions or give me a link to a site please. I really need help.

 

[Tide]

1. The probability that neither card is the red joker nor a red ace is $\frac {51}{54} \times \frac {50}{53}$ because there are 51 cards that are not a red joker or a red ace. Without replacement, there is 1 less card to choose from for the second draw.

Subtract that from 1 to find the probability that at least one of the two cards is the red joker or a red ace.

 

[HallsofIvy]

I have lots of question on probability because my teacher makes us learn from the book so i need help.
My first question is:
1. Suppose the two joker cards are left in a standard deck of cards. One of the jokers is red and the other is black. A single card is drawn is from the deck of 54 cards,but not returned, and then a second card is drawn. Determine the probability of drawing the red joker or a red ace on either draw.
This is the answer my teacher told me but i have no idea how to get 51/54 or the 50/53 so can anyone please explain this to me.
1- (51/54)*(50/53)= 1- 425/477 = 52/477

Tide explained this one.

[/quote]2. A basketball player has a success rate of 80% for shooting free throws.
Calculate the following probabilities.
i) She will make 3 out of 5 attempts[/quote]
Well, if you reall don't know how to do permutations, you could try this:(it's really doing the permutations "by hand".)
(Writing "S" for success and "F" for failure.)
She could get SSSFF which has probability (.80)(.80)(.80)(.20)(.20)= 0.02048.
She could get SSFSF which has probability (.80)(.80)(.20)(.80)(.20)= 0.02048.
She could get SSFFS which has probability (.80)(.80)(.20)(.20)(.80)= 0.02048
She could get SFSFS which has probability 0.02048- you should have seen by now that just changing the order of the numbers doesn't change the product.
She could get FSFSS which has probability 0.02048.
She could get FFSSS which has probability 0.02048.
She could get FSSFS which has probability 0.02048.
She could get FSSSF which has probability 0.02048.
She could get SFSSF which has probability 0.02048.
Now add them all up: since there were 10 different orders, that is 0.02048 times 10= 0.2048.
If you DID know "permutations" and could calculate that there are 5!/2!3!= 10 ways of writing the letters FFSSS, it would have been quicker.

ii)She will miss all 5

This is much easier: the probability of FFFFF is (.2)(0.2)(0.2)(0.2)(0.2)= (0.2)⁵= 0.00032.

iii)She will make the first 3 and miss the next 2.

Again, easier because there is no "permutation". The probability of
SSSFF is just (0.8)(0.8)(0.8)(0.2)(0.2) which, as we saw above, is equal to 0.02048.

This one i have no idea how to do because we are still on condition probability so i don't kno how to do permutations or anything
My last question is this:
An airplane can make a safe landing if at least half of its engines are working. Suppose that engine failures are independant events. Determine whether a two-engine plane is safer that a four engine plane if the chance that an engine fails is 1 in 2.

I want to go on record as saying that I wouldn't want to fly in an airplane which had engines having 1 chance in 2 of failing no matter how many engines it had!

A two engine airplane can make a safe landing if it has 1 or 2 engines still working. If the probability that an engine works is 0.5, then the probability that it has both engines working is (0.5)(0.5)= 0.25. The probability that it has exactly one engine working is 2(0.5)(0.5)= 0.50 (the two is because the probability engine 1 is working and engine 2 is not is 0.25 and the probability that engine 1 is not working and engine 2 is working is also 0.25: permutations again.) The probability that at least one engine is working is 0.25+ 0.50= 0.75.

A 4 engine airplane can make a safe landing if it has 2 or 3 or 4 engines working. The probability that all 4 engines work is (0.5)⁴= 0.0625. There are 4 ways to permute "WWWN" ("W"= working, "N"= not working) and each of those ways has probability (0.5)⁴= 0.0625 so the probability that exactly 3 engines is 4(0.0625)= 0.25.
There are 6 ways to permute "WWNN" and the probability of each of those is still (0.5)⁴= 0.0625 so the probability that exactly 2 engines are working is 5(0.0625)= 0.375,
The probability that at least 2 engines are working is the sum of those:
0.0625+ 0.25+ 0.375= 0.6875.

Looks like we are safer in the 2 engine airplane!

i also have no idea how to do this questions either.
So please can anyone explain step by step on how to do these questions or give me a link to a site please. I really need help.

 

[F.B]

O i forgot to post the last question to 13 which is probably harder than the other 3 so can anyone help me with this.

iv) She will make at least 3 out of 5