We know that Cubes with no surface painted can be find using , where x is number of cuttings. Here x=6.
If 1 is adjacent to 2, 3 and 5, then either 4 or 6 lies opposite to 1. So, the numbers 4 and 6 cannot lie opposite to each other. Hence, 4 necessarily lies adjacent to 6.
In a usual dice, the sum of the numbers on any two opposite faces is always 7. Thus, 1 is opposite 6, 2 is opposite 5 and 3 is opposite 4.
Consequently, when 4, 3, 1 and 5 are the numbers on the top faces, then 3, 4, 6 and 2 respectively are the numbers on the face touching the ground. The total of these numbers = 3 + 4 + 6 + 2 = 15.
If 1 is adjacent to 2, 4 and 6 then either 3 or 5 lies opposite to 1. So, the numbers 3 and 5 cannot lie opposite to each other. Hence, 3 is adjacent to 5 (necessarily).
We know that:
When one card is drawn from a pack of 52 cards
The numbers of possible outcomes n(s) = 52
We know that there are 26 red cards in the pack of 52 cards
? The numbers of favorable outcomes n(E) = 26
Probability of occurrence of an event: P(E)=Number of favorable outcomes/Numeber of possible outcomes=n(E)/n(S)
? required probability = 26/52 = 1/2.
Probability of occurrence of an event,
P(E) = Number of favorable outcomes/Numeber of possible outcomes = n(E)/n(S)
? Probability of getting head in one coin = ½,
? Probability of not getting head in one coin = 1- ½ = ½,
Hence,
All the 11 tosses are independent of each other.
? Required probability of getting only 2 times heads =
If 1 is opposite to 5 and 2 is opposite to 3, then 4 definitely lies opposite to 6. Therefore, 2 cannot lie opposite to any of the two numbers - 4 or 6. Hence, 2 necessarily lies adjacent to both 4 and 6.
Draw out the diagram and you will see that the width of the picture becomes 8.5 - 3 = 5.5
The length becomes 10 - 3 = 7. The area is now 5.5 x 7 = 38.5
Let the quantity of the wine in the cask originally be x litres
Then, quantity of wine left in cask after 4 operations =
litres
Suppose the can initially contains 7x and 5x litres of mixtures A and B respectively
Quantity of A in mixture left = litres.
Quantity of B in mixture left = litres.
So, the can contained 21 litres of A.
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