clicked photos of deer; 14 clicked photos of elephants and deer, 12 clicked photos of tigers and elephants, 15 clicked photos of deer and tigers, and 8 clicked photos of all the three animals. Each tourist took at least one photograph.
1. How many tourists went to the wildlife safari?
2. How many tourists clicked photographs of tigers only?
1. 43 venn diagram
2. 10 using venn diagram
done it roughly chances of having wrong are their..thank u.
NON
I would design a voice integrated clock that reads out the time aloud whenever people ask for time
Out of 10 persons, 4 are graduates; so, (10 – 4) = 6 are under-graduates.
If there is no restriction, any three can be chosen from the ten in (10C3) = 120 ways.
Now, if all three chosen are under-graduates; it can take place in (6C3) = 20 ways.
Therefore, the probability that there will be no graduate among the three chosen = (20 / 120) = (1 / 6).
Therefore, the probability that there will be at least one graduate among the three chosen = {1 – (1 / 6)} = (5 / 6) = 0.8333.
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9 sec=4 times
x sec=12 times
x=(9*12)/4=27 sec
( C ) Crawl
In a cube all the diagonal and sides are equal, we can go diagonally.
sow
Here is the solution to the given version of the puzzle (9 balls, one is heavier, need to identify oddball), where we label the balls A, B, …, I:
1. Weigh ABC versus DEF.
Scenario a: If these (1) balance, then we know the oddball is one of G, H, I.
2. Weigh G versus H.
Scenario a.i: If these (2) balance, the oddball is I.
Scenario a.ii: If these (2) do not balance, the heavier one is the oddball.
Scenario b: If these (1) do not balance, then the oddball is on the heavier side. For simplicity, assume the ABC side is heavier, so the oddball is one of A, B, C.
2. Weigh A versus B.
Scenario b.i: If these (2) balance, the oddball is C.
Scenario b.ii: If these (2) do not balance, the heavier one is the oddball.
answer is maximum of 2.