Some E maths TYS questions

bonkysleuth SG

Got some questions wrong. I looked at my methods at doing them again but I still can't pick out the mistakes.

There are 4 groups of cards in a pack. Two groups are red, one group blue and one group black. Each group has 5 cards labelled 1, 2, 3, 4 and 5.

Two cards are chosen at random, without replacement, from the pack

Find the probability that the cards chosen are

a red card labelled 1 and a blue card.

My solution: 2/20 X 5/19 = 1/ 38. The answer booklet is 1/19.

2 more questions to go. Hang on. I have to draw out the diagrams. (:

wee_ws SG

My solution: 2/20 X 5/19 = 1/ 38. The answer booklet is 1/19.

Hi,

P(red and blue) = P(red, blue) + P(blue, red) = 2 (1/38).

Thanks!

Cheers,
Wen Shih

Forbiddensinner SG

Hi,

P(red and blue) = P(red, blue) + P(blue, red) = 2 (1/38).

Thanks!

Cheers,
Wen Shih

In simple words, TS, you only calculated the possibility of getting a red card with 1 FIRST, followed by a blue card. You forgotten about the possibility of getting a blue card first, then getting a red card with 1.