Mathematics (1 Viewer)

[rick7]

Can someone please help me with these questions with working out?? The answers are there

Probability

1) a bag contains 3 yellow balls, 4 pink balls and 2 black balls. If the two balls are chosen at random, find the probability of getting a yellow and black ball.
a) with replacement
b) without replacement

answer: a) 4/27 b)1/6

2) Alan buys 4 tickets in a raffle in which 100 tickets are sold altogether. There are two prizes in a raffle. Find the probability that Alan will win
a) first prize
b) both prizes
c) 1 prize
d) no prize
e) attest 1 prize

answer:
a) 1/25
b)1/825
c)64/825
d) 152/165
e)13/165

3) Mary buys 20 tickets in a lottery that has 5000 tickets altogether. Find the probability that Mary will won
a) first and second prize
b) second prize only
c) neither first or second

answer:a) 19/ 1249750 b) 498/124975 c) 1239771/1249750

4) The two machines in a work shop each have a probability of 1/45 of breaking down. Find the probability that at any one time
a) neither machine ill be broken down
b) 1 machine will be broken down

5) A bag contains 5 black and 7 white marbles. Two are chosen at random from the bag
a) with replacement
b)without replacement
Find the probability of getting a black and a white marble

answer: a) 35/72 b)35/36

6) A card is chosen at random from a set of 10 cards numbered 1 to 10. A second card is chosen from a set of 20 cards numbered 1 to 20. Find the probability that the combination number these cards make is
a) 911
b)less than 100
c) between 300 and 500

Answer: a) 1/200 b) 81/200 c) 11/100

 

[parad0xica]

Can someone please help me with these questions with working out?? The answers are there

Probability

1) a bag contains 3 yellow balls, 4 pink balls and 2 black balls. If the two balls are chosen at random, find the probability of getting a yellow and black ball.
a) with replacement
b) without replacement

answer: a) 4/27 b)1/6

Amount of balls = 9.

Q1)

a) Pick a yellow ball (3/9), replace, then pick a black ball (2/9). Probability is 3/9 * 2/9 = 2/27.

OR pick a black ball first (2/9), replace, then pick a yellow ball (3/9). Probability is 2/27.

Add both to get 4/27.

b) Yellow first (3/9), do not replace, then pick a black (2/8). Probability is 3/9 * 2/8 = 1/12.

OR pick black first (2/9), do no replace, then pick a yellow (3/8). Probability is 2/9 * 3/8 = 1/12.

Add both to get 1/6.

_______________________

Note: Q1 and Q5 have the same idea.

 

[parad0xica]

3) Mary buys 20 tickets in a lottery that has 5000 tickets altogether. Find the probability that Mary will won
a) first and second prize
b) second prize only
c) neither first or second

answer:a) 19/ 1249750 b) 498/124975 c) 1239771/1249750

Q3)

a) Probability of winning first prize is 20/5000. Now Mary has 19 non-void tickets and the lottery has 4999 non-void tickets.

So probability of winning second prize is 19/4999.

Therefore probability of winning first & second prize is 20/5000 * 19/4999 = 19/1249750.

b) To win second prize only, she must not win first prize. P(not winning first prize) = 1 - P(winning first prize) = 1 - 20/5000. Now since she hasn't won any prizes, it means all her tickets are non-void, so she still has 20 tickets but lottery has 4999 non-void tickets.

So probability of winning second prize is 20/4999.

Therefore probability of winning second prize only is (1 - 20/5000) * (20/4999) = 498/124975.

c) Mary must not get first and second prize.

P(not winning first prize) = 1 - 20/5000 (as above)

P(not winning second prize at this point) = 1 - P(winning second prize at this point) = 1 - 20/4999.

So probability of winning neither first or second is (1 - 20/5000) * (1 - 20/4999) = 1239771/1249750.

______________

Note: Q2 and Q3 have the same idea. I did the easier one so you can build on fundamentals.

 

[parad0xica]

6) A card is chosen at random from a set of 10 cards numbered 1 to 10. A second card is chosen from a set of 20 cards numbered 1 to 20. Find the probability that the combination number these cards make is
a) 911
b)less than 100
c) between 300 and 500

Answer: a) 1/200 b) 81/200 c) 11/100

Q6) Structure of combination is: _ _ _

a) Pick a 9, then pick an 11

P(pick a 9 from first set) = 1/10.

P(pick an 11 from second set) = 1/20.

Therefore, P("911") = 1/10 * 1/20 = 1/200.

b) Cannot pick 10 otherwise we get a number greater than 100. So pick 1 or 2 or ... or 9 then pick 1 or 2 or ... or 9 to generate all 2-digit numbers less than 100.

I don't know a better way to do it within 2U bounds but here is one long way...

Wait, maybe P(less than 100) = 1 - P(greater than 100) could be faster and less complex

Case 1:

Sub-case 1.1: pick 1 first, then pick 1. Probability = 1/10 * 1/20.
Sub-case 1.2: pick 1 first, then pick 2. Probability = 1/10 * 1/20.

.
.
.

Keep going, there are 9 cases and 9 sub-cases for each meaning 81 possibilities. So probability is 81 * 1/10 * 1/20 = 81/200.

c) First card must be either 3 or 4, then second card must be a 2-digit number (10, 11, ..., 20), eleven of them.

Case 1: pick 3, then pick a 2-digit number. Probability = 1/10 * 11/20 = 11/200.

Case 2: pick 4, then pick a 2-digit number. Probability = 1/10 * 11/20 = 11/200.

Just add the probabilities,

P(case 1 OR case 2) = 11/200 + 11/200 = 11/100.

 

[parad0xica]

4) The two machines in a work shop each have a probability of 1/45 of breaking down. Find the probability that at any one time
a) neither machine will be broken down
b) 1 machine will be broken down

Q4)

P(1 machine break down) = 1/45.

P(1 machine not break down) = 1 - 1/45 = 44/45.

a) P(neither break down) = 44/45 * 44/45.

b) P(1 machine break down AND 1 machine no break down(?)) = 1/45 * 44/45.

 

[rick7]

Q3)

a) Probability of winning first prize is 20/5000. Now Mary has 19 non-void tickets and the lottery has 4999 non-void tickets.

So probability of winning second prize is 19/4999.

Therefore probability of winning first & second prize is 20/5000 * 19/4999 = 19/1249750.

b) To win second prize only, she must not win first prize. P(not winning first prize) = 1 - P(winning first prize) = 1 - 20/5000. Now since she hasn't won any prizes, it means all her tickets are non-void, so she still has 20 tickets but lottery has 4999 non-void tickets.

So probability of winning second prize is 20/4999.

Therefore probability of winning second prize only is (1 - 20/5000) * (20/4999) = 498/124975.

c) Mary must not get first and second prize.

P(not winning first prize) = 1 - 20/5000 (as above)

P(not winning second prize at this point) = 1 - P(winning second prize at this point) = 1 - 20/4999.

So probability of winning neither first or second is (1 - 20/5000) * (1 - 20/4999) = 1239771/1249750.

______________

Note: Q2 and Q3 have the same idea. I did the easier one so you can build on fundamentals.

thank you for your help
but can u explain q2 though?

 

[davidgoes4wce]

Draw a tree diagram to help with you in Question 2:

 

[dan964]

Q6) Structure of combination is: _ _ _

a) Pick a 9, then pick an 11

P(pick a 9 from first set) = 1/10.

P(pick an 11 from second set) = 1/20.

Therefore, P("911") = 1/10 * 1/20 = 1/200.

b) Cannot pick 10 otherwise we get a number greater than 100. So pick 1 or 2 or ... or 9 then pick 1 or 2 or ... or 9 to generate all 2-digit numbers less than 100.

I don't know a better way to do it within 2U bounds but here is one long way...

Wait, maybe P(less than 100) = 1 - P(greater than 100) could be faster and less complex

Case 1:

Sub-case 1.1: pick 1 first, then pick 1. Probability = 1/10 * 1/20.
Sub-case 1.2: pick 1 first, then pick 2. Probability = 1/10 * 1/20.

.
.
.

Keep going, there are 9 cases and 9 sub-cases for each meaning 81 possibilities. So probability is 81 * 1/10 * 1/20 = 81/200.

c) First card must be either 3 or 4, then second card must be a 2-digit number (10, 11, ..., 20), eleven of them.

Case 1: pick 3, then pick a 2-digit number. Probability = 1/10 * 11/20 = 11/200.

Case 2: pick 4, then pick a 2-digit number. Probability = 1/10 * 11/20 = 11/200.

Just add the probabilities,

P(case 1 OR case 2) = 11/200 + 11/200 = 11/100.

For part (b)
you can simplify further

the requirement is for a number to be less than 100, both numbers must be of 1, 2, 3, 4, 5...9 = 9 possible numbers that will do.

For part (c):
One can pick 3,4 as the leading digit:
2/10

One can pick 10,11,...20 (11 possible numbers)
=2/10*11/20
=22/200 as above.

 

[parad0xica]

For part (b)
you can simplify further

the requirement is for a number to be less than 100, both numbers must be of 1, 2, 3, 4, 5...9 = 9 possible numbers that will do.

For part (c):
One can pick 3,4 as the leading digit:
2/10

One can pick 10,11,...20 (11 possible numbers)
=2/10*11/20
=22/200 as above.

Thanks. I don't know why I didn't use that fundamental probability concept in this question even though I used it in Q3