Probability

[Rone123]

Suppose we have 3 cards identical in form except that both sides of the first card are colored red, both sides of the second card are colored black, and one side of the third card is colored red and the other side is colored black.
The 3 cards are mixed up in a hat, and 1 card is randomly selected and put down on the ground. If the upper side of the chosen card is colored red, what is the probability that the other side is colored black?

[Thokkalee]

[DalchanChekka]

1/2

[Rone123]

think about it. Answer is 1/3. 

if you are interested, read Baye’s theorem on conditional probabilities and Monty Hall game

[Rone123]

..

[hunkyfunky2]

1/2

[Rone123]

41 minutes ago, hunkyfunky2 said:

1/2

Answer is 1/3.

out of total 6 sides, 3 are colored red. Let’s mark the red sides as A,B (two sides of same card), C other card.

 1. A is up, B is down

2. B is up, A is down

3. C is up, and down is black

 

[Rone123]

Hope this clarified it,

[Starblazer]

There are 6 sides of all 3 cards, among them 3 are reds and 3 are blacks.

Only 2 cards have at least one side red. They have 4 sides with 3 red and 1 black.

If the upper side is red, then we have lower side either (remaining) 2 reds or 1 black. The probability of getting black is 1/3.

i.e., 1 black / (2 reds+1 black).

[Rone123]

4 hours ago, Starblazer said:

There are 6 sides of all 3 cards, among them 3 are reds and 3 are blacks.

Only 2 cards have at least one side red. They have 4 sides with 3 red and 1 black.

If the upper side is red, then we have lower side either (remaining) 2 reds or 1 black. The probability of getting black is 1/3.

i.e., 1 black / (2 reds+1 black).

Correct 👏 

[texas]

8 hours ago, Rone123 said:

Suppose we have 3 cards identical in form except that both sides of the first card are colored red, both sides of the second card are colored black, and one side of the third card is colored red and the other side is colored black.
The 3 cards are mixed up in a hat, and 1 card is randomly selected and put down on the ground. If the upper side of the chosen card is colored red, what is the probability that the other side is colored black?

1/4

[Swatkit]

9 hours ago, DalchanChekka said:

1/2

anna vendy ki eedio call cheshe answer gistunde ga @Swatkat @anna_vendy @Swatkit

simple math swatkat annaoi parade vachela istunde ga 

[Swatkit]

9 hours ago, Rone123 said:

Suppose we have 3 cards identical in form except that both sides of the first card are colored red, both sides of the second card are colored black, and one side of the third card is colored red and the other side is colored black.
The 3 cards are mixed up in a hat, and 1 card is randomly selected and put down on the ground. If the upper side of the chosen card is colored red, what is the probability that the other side is colored black?

1/3 ani swatkat anan cheppamnadu anna vendy eyyi 

[Ministryofbadmouth]

8 hours ago, Rone123 said:

Answer is 1/3.

out of total 6 sides, 3 are colored red. Let’s mark the red sides as A,B (two sides of same card), C other card.

 1. A is up, B is down

2. B is up, A is down

3. C is up, and down is black

 

there is one card in 3 cards which can show red up and down black...its straight forward picking up any one card out of 3 is 33.335 which is 1/3..

dont make it complex..when solution is simple

[Rone123]

Yes, it works out that way. But not obvious, antha easy ayi vunte, Bayes theorem ki aa name vachedi kadu kada