Counting - PnC , Probability

RE: Counting

aniket prajapati — Mon, 15 Oct 2018 15:05:48 +0000

The center of the coin is at least one coin radius from any grid line.
You can then shade the area of a grid square where the coin center cannot fall -- this shaded area will look the same in every square. If the coin radius is r and the grid square sides have length a, there's a small square of side length a−2r in each square where the center can fall without the coin extending beyond the grid square.

So the probablity of staying within the square becomes

(a-2r)²/a²

putting the given values probability=64/100=16/25

New Question: Counting

Ridhima Mehta — Mon, 15 Oct 2018 14:21:23 +0000

404

any other approach to this TG solved example? Unable to comprehend the given solution.

New Question: Counting

Ridhima Mehta — Mon, 15 Oct 2018 14:19:46 +0000

403

solution please

RE: Counting

aniket prajapati — Fri, 05 Oct 2018 16:54:47 +0000

Alternate approach,

If the new colour doesn't contain Blue paint, then it isn't of type Jaune. So, it is much easier to calculate the complementary probability.
P(no Blue paint chosen) = (5/9)*(4/8)*(3/7) = 5/42.
Therefore, the required probability is 1 - 5/42 = 37/42

RE: Counting

aniket prajapati — Fri, 05 Oct 2018 15:07:22 +0000

Jaune Y = (4 C1)*(5 C 2) = 4*10 = 40
Jaune X = (4 C 2)*(5 C 1) + (4 C 3) = 6*5 + 4 = 34
Total combinations = 9C 3 = 84

Probability of Jaune = (40 + 34)/84 = 37/42

RE: Counting

aniket prajapati — Fri, 05 Oct 2018 15:03:10 +0000

if we assume that each couples as a one person total 3 person in a row and the probablity of three person, where couples seat together, on five chairs is 3/5.

But the question is asking neither of couples seat together so the opposite way 1-3/5=2/5

New Question: Counting

Richa — Fri, 05 Oct 2018 11:07:53 +0000

333
Sir, please help.

New Question: Counting

Richa — Thu, 04 Oct 2018 18:29:17 +0000

332

RE: Counting

aniket prajapati — Thu, 04 Oct 2018 10:15:04 +0000

330 is it right sir ?

RE: Counting

TG Team — Tue, 02 Oct 2018 19:39:57 +0000

Divisible by 12 means divisible by 3 and 4 both.

to be divisible by 4 last two digits has to be divisble by 4. only combination : 76

_ _ _ _ _ _ _ 76

to be divisble by 3 sum of the digits has to be divisible by 3 .

sum of last two digits = 7 + 6 i.e 13 ( 3k + 1)

so sum of remaining 7 digits should be of the form 3k + 2 .
_ _ _ _ _ _ _
All sevens => sum 49 ( 3k + 1) hence not divisble by 3 .

6 sevens , 1 sixes => sum 48 ( 3k ) not divisble by 3.

5 sevens , 2 sixes => sum 47 ( 3k + 2) divisble by 3.

7777766 => 7!/(5!2!) numbers

4 sevens , 3 sixes => not divisble by 3

3 sevens , 4 sixes => not divisble by 3

2 sevens , 5 sixes => divisble by 3.
6666677 => 7!/(5!2!) numbers

1 seven , 6 sixes => not divisble by 3
all sixes => not divisible by 3

Hence , 2 x 7!/(5!2!) = 42 , 9 digit numbers