Probability tree help plz (1 Viewer)

[chooky_girl26]

can someone plz help me with theses questions... i don't understand them at all... can't make out how they got the answers in the back of the book.

Jia is a shooter with an 80% change of hitting a target. If he has three shots at a target, find the probability that:
a. He hits with all three shots
b. He hits with exactly two shots

A raffle has 100 tickets with two prises. Kevin buys five tickes. Find the probability that:
a. Kevin wins 1st prize= 5/100 = 1/20 (i can do that much)
b. Kevin wins both prizes ( the back of the book says 1/495 how do u get that?)
c. Kevin does not win a prize (another odd answer that is unexplainable)
d. Kevin wins exactly one prize S HELLPPP)

6. A bag contain four black and six white marbles. Two marbles are drawn from the bag one after the other. If the first marble drawn is black, the probibility that the second marble drawn is white is :

A coin is biased such that the probability of it landing heads is 0.6. The coin is tosed three times. Which of the following outcomes has the greatest probability of occuring?
Tossing three heads
Tossing two heads and one tail
Tossing one head and two tails
Tossing three tails


Sorry about so many questions.... but i am really struggling with this... i need a little help.. if u could help me.. it'd be great thanks all
Mwa
xoxo

 

[PC]

The best thing to do with these questions is to draw a nice clear probabiity tree and write the probability on the branches. [See attachment]

In your first question, the probability is a percentage, which represents an average chance. This will apply to all branches.

In your second question, you need to look out for "non-replacement". There are 100 tickets. When the first tickets is drawn, then only 99 tickets remain. If Kevin wins the first prize then he only has 4 tickets remaining in the draw for second prize, and so on.

Have a look at these answers and see how you go with the other questions.

Hope this helps

 

[Trev]

6. A bag contain four black and six white marbles. Two marbles are drawn from the bag one after the other. If the first marble drawn is black, the probibility that the second marble drawn is white is :

A coin is biased such that the probability of it landing heads is 0.6. The coin is tosed three times. Which of the following outcomes has the greatest probability of occuring?
Tossing three heads
Tossing two heads and one tail
Tossing one head and two tails
Tossing three tails

Probability second marble drawn white is:
4B, 6W; since one black taken out already there is 3B marbles. Therefore total of (3+6) left, 3 being white. P(2nd White) = 3/6 = 1/2.

P(HHH) = 0.6*0.6*0.6 = 0.216
By inspection it is clearly 3 heads that has the greatest probability of occuring; however I will continue;
P(HHT) = 0.6*0.6*0.4 = 0.144
P(HTT) = 0.6*0.4*0.4 = 0.096
P(TTT) = 0.064

P(HHH) > P(HHT) > P(HTT) > P(TTT)
So, P(HHH), or 3 heads, has the greatest probability of occuring.

 

[chooky_girl26]

thanks guys

thanks guyss u've helped me alot... but i still got loads and loads of questions... i'm so confused... could someone plz add me to msn if u have it?? miss_lysha@hotmail.com... coz i need so much help with this probability stuff "( thanks

 

[Trev]

Add me if you like, party_guy_007@hotmail.com.
I tried adding you, but it didn't work!

 

[chooky_girl26]

expected outcomes

Someone come help me with this question plz

Two dice are rolled 100 times Copy and complete the table below to caculate the expected number of occurrences of each total in 100 rolls of dice. Give each answer correct to 1 decimal place.


Outcome 2 3 4 5 6 7 8 9 10 11 12

probability

Expected no.

Someone plz help
Cheers

 

[PC]

Someone come help me with this question plz

Two dice are rolled 100 times Copy and complete the table below to caculate the expected number of occurrences of each total in 100 rolls of dice. Give each answer correct to 1 decimal place.


Outcome 2 3 4 5 6 7 8 9 10 11 12

probability

Expected no.

Someone plz help
Cheers

The best way to do this one is draw up a table. Put the numbers on one die along the top row, and the numbers of the other die down the left column. Then work out the total from each combination.

| 1 2 3 4 5 6
--------------------------
1 | 2 3 4 5 6 7
2 | 3 4 5 6 7 8
3 | 4 5 6 7 8 9
4 | 5 6 7 8 9 10
5 | 6 7 8 9 10 11
6 | 7 8 9 10 11 12

So there are 36 possible outcomes. There is only one way of getting a 2, two ways of getting a 3, and so on.

Probability of 2 = 1/36
Probability of 3 = 2/36 = 1/18
Probability of 4 = 3/36 = 1/12
Probability of 5 = 4/36 = 1/9
Probability of 6 = 5/36
Probability of 7 = 6/36 = 1/6
Probability of 8 = 5/36
Probability of 9 = 4/36 = 1/9
Probability of 10 = 3/36 = 1/12
Probability of 11 = 2/36 = 1/18
Probability of 12 = 1/36

To find the expected number, multiply the probability by the number of throws.

Number of 2s = 1/36 x 100 = 3
Number of 3s = 1/18 x 100 = 6
Number of 4s = 1/12 x 100 = 8
Number of 5s = 1/9 x 100 = 11
Number of 6s = 5/36 x 100 = 14
Number of 7s = 1/6 x 100 = 17
Number of 8s = 5/36 x 100 = 14
Number of 9s = 1/9 x 100 = 11
Number of 10s = 1/12 x 100 = 8
Number of 11s = 1/19 x 100 = 6
Number of 12s = 1/36 x 100 = 3

Hope this helps.

 

[chooky_girl26]

i'm still confused.... do u have msn?? if so could u plz add me

i don't get how u get the numbers