16 rotten bananas are accidentally mixed with 155 good ones. It is not possible to just look at a banana and tell whether or not it is rotten. One banana is taken out at random from this lot. Determine the probability that the banana is taken out is a good one.
Numbers of bananas = Numbers of rotten bananas + Numbers of good bananas ∴ Total number of bananas = 155 + 16 = 171 bananas P(E) = (Number of favourable outcomes) / (Total number of outcomes) P(picking a good banana) = 155/171
2(1/3) + 2(5/6) – 1(1/2) = ? – 6(1/6)
63- [22-{24 ÷ 3-(9-15 ÷ 5) ÷ 6}]=?
1000÷ 250 = ( 3√? × √1444) ÷ ( 3√512 × √361)
(〖(0.4)〗^(1/3) × 〖(1/64)〗^(1/4) × 〖16〗^(1/6) × 〖(0.256)〗^(2/3))/(〖(0.16)〗^(2/3) × 4^(-1/2) ×〖1024〗^(-1/4) ) = ?
(3500 ÷ √1225) × √(20.25) = ? ÷ 4
√ 225 x 24 - √ 144 x 18 = ?
[123 ÷ 8 ÷ 9] × 144 = ? + 12 × 5
[(36 × 15 ÷ 96 + 19 ÷ 8) × 38] = ?% of 608
2/5 of 3/4 of 7/9 of 14400 = ?
(√196 + √121) × 4 = ?/2