Question
22 rotten bananas are accidentally mixed with 136 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.Solution
Numbers of bananas = Numbers of rotten bananas + Numbers of good bananas ∴ Total number of bananas = 136 + 22 = 158 bananas P(E) = (Number of favourable outcomes) / (Total number of outcomes) P(picking a good banana) = 136/158 = 68/79
?% of (112.31 ÷ 13.97 × 90.011) = 359.98
40.22 of 249.98% + 459.99 ÷ 23.18 = ?
456.9 + 328.10 - 122.98 = ? + 232.11
108.31% of (4.9/9.012) of ? = 23.9% of 2499.9
A man lost one-fourth of his initial amount in the gambling after playing three rounds. The rule of Gambling is that if he wins he will receive Rs. 1000...
What approximate value will come in place of the question mark (?) in the following question? (Note: You are not expected to calculate the exact value.)...
(32.18% of 2399.89 - √624 × 26.25) % of 149.79 = ?
1299.99 ÷ 20.21 = ? + 325.985 - (180 ÷ 6 × 24.03)
256.12 ÷ 7.92 + 26.11 × 7.82 – 44.09 = ?2
