To be unable to solve this maths questions????

[Brokensoul]

1). Kit picked one card at random from a pack of 52. Write down the probability that he will pick : a) a ten
b) a red card
c) a heart
d) the ace of diamonds?

2)There are three toffees and one mint left in a bag of sweets. Ben says that the probability of picking the mint is 1/3 because there are three toffees and only one mint. Is he correct and why?

Am I dum dum that I am confused ? Please help! Thank you.

1a) 4/52 as there are 4 tens and 52 cards so the chance of getting a ten is 4 out of 52
b) 13/52
c)13/52
d)1/52

2 No he is not correct because there are 4 sweets altogether so he has a one in four chance of picking a mint

26/52 for a red card missmapp!

Ten 4/52, red 26/52, heart 13/52, ace of d 1/52,

No, prob 1/4 as four sweets 1 mint, prob of toffee is 3/4,
Hope thats right and helps on phone

Draw pictures then. That will explain it in a visual way

Oh yes, bugger!! it is Friday

Basically the number of the given option( red card, toffee etc) is out of the total number of things ( cards , sweets etc)

The fraction shows the probability of picking an option OUT OF a larger group

I'd think there is 1 in 4 chance of picking a 10, 1 in 24 of a red card, 1 in 12 of a heart and 1 in 52 of the ace of diamonds.

No 1/3 not correct as 4 sweets to.choose from so it's a 1 in 4 chance.

1 in 13
1 in 2
1 in 4
1 in 52

1 in 4

"So you need to work out how many red cards there are in a pack."

Yes, that's right. E.g. there are 4 tens in a pack, so there are 4 chances to pick a ten out of 52 cards, so the probability is 4/52 (which can be simplified down to 1/13 if you prefer). There are 26 red cards so it's 26/52, which simplifies down to 1/2 which makes sense as half the pack are red.

Thank you soooooo much. My dd has got so much homework over the weekend that I wanted to help. My brain just stopped. THANK YOU SOOOO MUCH... I am really thankful.

1 a - 4 in 52
1b - 26/52 or 1 in 2
1 c - 1 in 4 or 13/52
1d - 1 in 52

2 1 in 4

The question hasn't said it's a standard pack of playing cards, it's just said a pack of 52. It might've been 52 business cards, so zero is a good answer for all of the first ones.

Red card is 1 in 26 and heart 1 in 13!

Why did I think there were 12 in each suit not.13!

You are all amazing. I wouldn't be able to do that and that's why I asked you. It's hard for me.... Thank you so much. I have more q but maybe later. Thank you soooo much

CatsRule

Red card is 1 in 26 and heart 1 in 13!

Nope, red card is 1 in 2 and heart is 1 in 4.

Could the answer to the bag of sweets one alternately be 1 in 2? You will either pick a mint or not. It's like a well known problem about which door to pick, which I can't recall just now.

Some worrying maths ideas on here!

RooRooTaToot

Could the answer to the bag of sweets one alternately be 1 in 2? You will either pick a mint or not. It's like a well known problem about which door to pick, which I can't recall just now.

Probability fail.

The door scenario is usually quoted as a game show where the contestant chooses one of three doors in the hope of winning a prize which is behind one of the doors. The host then opens one of the doors that wasn't picked and doesn't have the prize behind it.

This leaves two unopened doors, with the prize still behind one of them. The host gives the contestant the opportunity to change doors if they wish. Is it wise to keep the same door, or swap to the other door?

The bag if sweets would only be 1 in 2 if there was an equal chance of any one sweet being a mint. Which there is not, because only one of the 4 is a mint

The doors thing is I think like the question of flipping a coin - question is usually something like if you have flipped heads ten times in a row, what are the chances the next will be heads? The answer is 1/2 just like every other time. The probability of getting 11 heads in a row however is lower than getting 10 and then a tails - that's the bit that gets confusing!

Ignore me, I was getting the principles of the Monty Hall problem messed up with something else in my head. I used to be so good at maths too. Need to refresh my skills I think.

mrsmalcolmreynolds

The probability of getting 11 heads in a row however is lower than getting 10 and then a tails - that's the bit that gets confusing!

I'm fairly sure (without working it out) that the probability is the same in both cases: 1 in 2^11.