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If a leap year is selected at random, what is the chance that it will contain 53 Tuesdays?
We know that, in a leap year there are total 366 days, 52 weeks and 2 days
Now, in 52 weeks there are total 52 Tuesdays
∴ Probability that the leap year will contain 53 Tuesdays is equal to the probability of remaining 2 days will be Tuesdays
Thus, the remaining two days can be:
(Monday and Tuesday), (Tuesday and Wednesday), (Wednesday and Thursday), (Thursday and Friday), (Friday and Saturday), (Saturday and Sunday) and (Sunday and Monday)
∴ Total Number of cases = 7
Cases in which Tuesday can come = 2
Hence, probability (leap year having 53 Tuesdays)
Suppose we have four boxes A, B, C and D containing coloured marbles as given below:
Box | Marble colour | ||
Red | White | Black | |
A | 1 | 6 | 3 |
B | 6 | 2 | 2 |
C | 8 | 1 | 1 |
D | 0 | 6 | 4 |
One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from box A?, box B?, box C?