2..3...5..7
3..5..7
5..7
5...7
7...11
7...11

total=15
15/36= 5/12

2-required observation

Sample space-- 1,2,3,4,5,6

Probability=1/6

Total black =26 out of 52
Frst card - black probability = 26/52 = 1/2
Left card totl = 51
Left black = 25
2nd blck probability = 25 /51
Total probability = 1/2*25/51 = 25/102

x+5y+10z=38

y=3
Z=2
X=3

Or

X=3
Z=3
Y=1


4c3*5c3*3c1/(12c7)
+
4c3*4c3*5c2/(12c8)

??

bag contains 5 red balls, 6 yellow balls and 3 green balls. If two balls
are picked at random, what is the probability that either both are red or both
are green in colour?

5C2 + 3C2/ 14C2

6c2/xc2=5/92
15/xc2=5/92
15/5=3
92*3=276
X(x-1)=276*2
X^2-x-552=0
24*23=552
24-18=6
10-8

Pls explain q 60-62

q.60

(4c2+5c2)/12c2

=16/66=8/33

61

8c3/12c3 =56/220 =14/55


62

1- 7c3/12c3 =37/44

4×3×2×1 |4

(12c10×12c8×12c9×12c11×12c12)×5fac
= 95280