Prime factors of 45 are 3 x 3 x 5 i.e. 3 and 5.
The smallest number to be multiplied by 45 so that it may have 3 distinct prime factors is 2.
Number become 90 (45x2) and factors are 2x3x3x5.
Probability of at least one attending the class = 1- probability of none attending
Let probability of A, B and C attending the class be a, b, c respectively.
Thus, probability of not attending the class by A, B and C is (1 - a), (1 - b) and (1 – c) respectively.
The probability of exactly one of A or B attending = a(1 - b) + b(1 - a) = 710
= a + b – 2ab = 710 (1)
Similarly, for B or C = b + c – 2bc = 410 (2)
And, for C or A = c + a – 2ca = 710 (3)
Probability of all three attending (i.e. abc) = 99100 (4)
From (1), (2) and (3): 2(a + b + c – ab – bc - ca) = 710 +410 + 710 = 1810 = 95
a + b + c – ab – bc – ca = 910
Probability of none attending = (1 - a)(1 - b)(1 – c) = 1 – a – b – c + ab + bc + ca – abc
Probability of none attending = 1 – (a + b + c – ab – bc – ca + abc)
Probability of none attending = 1 – (910 + 99100) = 1100
Probability of at least one attending = 1 – probability of none attending
Probability of at least one attending = 1 – 1100 = 99100
As any drawn fruit will be a banana so the probability of drawing a banana is 1
The shopkeeper sold two T-shirts for Rs. 2400
Selling Price of one T-shirt = Rs. 1200
Profit = 33.33%
Cost Price = S.P. * 100100+P% = 1200 * 100100+33.33 = 1200 * 100133.33 = 1200 * 34 ≈ Rs. 900
What is the value of the expression 4_{23} * 5_{20} * 6_{-2} * 3_{2} * 5_{-5} * 2_{-46} * 5_{-10} * 11_{0} * 5_{-5}?
4_{23}*5_{20}*3_{2}*11_{0}6_{2}*5_{5}*2_{46}*5_{10}*5_{5} = (2_{2})_{23}3*5_{20}*3_{2}*1(3*2)2*5_{5}+10+5*2_{46} = 2_{46}*5_{20}*3_{2}*12_{2}*3_{2}*5_{20}*2_{46} = 12_{2} = 14