i.e., P(E_{1 }∪ E_{2} ∪ E_{3} .... ∪ E_{n})

= P(E_{1}*) + * P(E_{2}) +.... + P(E_{n})

- Option : A
- Explanation :

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- Option : B
- Explanation : The five digits can be arranged in 5! ways, out

of which 4! will begin with zero.Total number of 5-figure numbers formed

= 5! - 4! = 96.

Those numbers formed will be divisible by 4

which will have two extreme right digits

divisible by 4,

i.e., numbers ending in 04, 12,20,24,32,40.

Now, numbers ending in 04 = 3! = 6,

numbers ending in 12 = 3! - 2! = 4,

numbers ending in 20 = 3! = 6,

numbers ending in 24 = 3! - 2! = 4,

numbers ending in 32 = 3! - 2! = 4,

and numbers ending in 40 = 3! = 6.

[Numbers having 12, 24, 32 in the extreme right

are (3! - 2!), since the numbers having zero on

t.he extreme left are to excluded.]

Total number of favourable ways

6 + 4 + 6 + 4 + 4 + 6 = 30

Hence, required probability =

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- Option : B
- Explanation : Two balls out 14 can be drawn in
^{14}C_{2}ways which is the total number of outcomes.Two white balls out of 8 can he drawn in

^{8}C_{2}waysProbability of drawing 2 white balls

Similarly 2 red balls out of 6 call be drawn in

^{6}C_{2 }ways

Probability of drawing 2 red balls

Probability of drawing 2 balls of the same

colour (either both white or both red)

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