Math Games

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Question: Math Probability - Dice and Cards?

The Aces and Kings are removed from a standard deck of playing cards and then the following little game is played. Pick a card from the well-shuffled, reduced deck and then roll a pair of dice.
Find the probability that:

a) the face value of the selected card matches that of the sum obtained on the dice (assume that Jacks and Queens have values of 11 and 12 respectively.)
b) the card value is 10 or the dice value is 10. Are the events here mutually excluive? Independent? Briefly Ex

Answer: a) You can do a long calculation adding up probabilities of both 2, both 3 etc but this is not necessary. Whatever the dice total comes to there will be a 1/11 probability that the card matches it.

b) Card value 10 and dice value 10 are not mutually exclusive because they can both occur at the same trial. They are independent because the outcome of one does not affect the probability of the other.

Dice Trick Solution

Question: I need help with these math problems?

1) Group the consecutive counting numbers as follows: (1), (2,3), (4,5,6), (7,8,9,10), . . Notice that there is one number in the first group, two numbers in the second group, three in the third, etc. What is the sum of the numbers in the 99th group?

2) Two students are playing a math game that uses sixteen different numbered tiles laying face down on a table. The faces of the tiles are labeled with one of the numbers from 1 to 16. If each student turns over one card, what is the probability that the sum of the faces of the two tiles is even?

3) The digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 are repeated in such a way that when the digits repeat the greatest digit is removed:

0123456789012345678012345670123456012345012340123012010
0123456789012345678012345670123456012345012340123012010……

The pattern is repeated indefinitely. What is the 1111st digit in this pattern?
Thank you so much

Answer: 1) The grouping
(1), (2, 3), (4, 5, 6), (7, 8, 9, 10),

Has n numbers in the nth group
And the last number in each group is as follows
1st group ---- 1
2nd group ---- 1+2 = 3 = (2*3)/2
3rd group -------1+2+3 = 6 = (3*4)/2
4 th group ----------1+2+3+4 = 10 or 4(4+1)/2
Nth group – last number is n*(n+1)/2

For 99th group last number is 99*100/2 = 4950

For 98th group last number is 98*99/2 = 4851

Now the sum of the numbers in 99th group is
Sum of numbers for 99 groups – sum of numbers in 98 groups
Number are in App with d =1, a= 1 or natural numbers
Hence (4950)*(4951)/2 – (4851)* (4852)/2
= 12253725 – 11768526 = 485199 (Ans)

2) I presume your tile and card is same
The possibility of having the total being even is
Both select an odd card or both select an even card.

The sample space is 16*16 = 256
(Each select one of the tiles from 16 tiles)

Now for each of the two to select an odd tile we have 8*8 events
And for each of the two to select an even tile we have 8*8 events
So successful event is 8*8 + 8*8 = 128

Hence probability is 128/256 = ½ (Ans)

3) Your question is not very clear
Which pattern repeats because after 1 set of numbers are exhausted we are let with no numbers
So we won’t have an 1111th no

But rephrasing you question with a n assumption that the entire arrangement repeats
Namely after we use (0-9) + (0-8) ……--------------- (0-1) and 0
We begin again with (0-9) + (0-8) ---------------------
And so on

We observe that repetition occurs after every [10 + 9 + 8 +…….1) = 10*11/2 = 55 no of items

(Here we have 10 numbers 0-9, because 0 is allowed)

Now we have 1111th terms
Hence we have 1111 = 55 * 20 + 11 (set of 55 repeated 20 times and 11 terms)

Now the 11th term is 0 for the above arrangement (Ans)

Math Games For Kids : Playing The Pattern Game