Question

there are 6 bag in each bag has 5 balls how many balls will one third of these bags have?​

Answer 1

Answer:

There are 5 black and 6 red balls in the bag.

2 black balls can be selected out of 5 black ball in 5C2 ways and 3 red ball can be selected out of 6 red balls in 6C3 ways.

Thus, by multiplication principle, required number of ways of selecting 2 black and 3 red balls is  5C2×6C3

=2!3!5!×3!3!6!

=25×4×3×2×16×5×4=10×20=200

Answer 2

Answer:

$\huge\mathbb\fcolorbox{purple}{lavenderblush}{✰Solution}$

Step-by-step explanation:

21 is the only number which is multiple of both 3 and 7, so while calculating probability separately for 3 and 7, this number 21 will be taken twice so to avoid confusion we take it once here and will exclude 21 while calculating probability separately. Also, number 21 can only be taken once.

➡ Probability of getting number multiple of both 3 and 7 (number 21) =1/28

➡ Probability of getting numbers which are multiple of 3 ( 3,6,9,12,15,18,24,27) = 9 / 28

➡ Probability of getting numbers which are multiple of 7 (7,14,28) =3/28

➡ Probability of getting a number either multiple of 3 or 7 = 3/28 + 9/28 + 1/ 28 = 12/28 = 3/7

Hence 3/7 (a) is the right answer.

Hope it helps you..