A super lotto has all the two digit numbers written on balls kept in

Question

A super lotto has all the two digit numbers written on balls kept in a bag. If two balls are chosen at random without replacement what is the probability that the two numbers have the same units digit and the product of the two numbers has the same units digit as the two numbers?

A.  \frac{2}{5}

B.  \frac{12}{445}

C.  \frac{16}{445}

D.  \frac{8}{445}

E.  \frac{2}{25}

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Nidzo · Answer

Number of 2 digit numbers: 90

Unit numbers which when multiplied together yield a number with the same unit digit: 0, 1, 5 and 6. Each of these units will appear a total of 9 times between 10 and 99.

In total there are 36 numbers which end in either 0, 1, 5 or 6.

The probability that the first number ends in either 0, 1, 5 or 6:   \frac{36}{90}  or  \frac{2}{5}

The probability of drawing a number with the same units digit:  As there are 9 in total for each, and one has already been drawn, there are 8 remaining. 
So the probability will be  \frac{8}{89}

Probability that the two numbers will end with the same units digit, which when multiplied gives the same unit's digit:

\frac{2}{5}*\frac{8}{89}

\frac{16}{445}

Answer C

IanStewart · Answer

We need to pick two numbers that end in 0, 1, 5 or 6, since those are the only units digits which, when squared, don't change. If we were picking with replacement, we'd have a 4/10 chance of picking a number with one of those units digits, then a 1/10 chance of picking a number with a matching units digit, and the answer would be 4/100. When we're picking from a set this large, picking without replacement will barely change our probabilities, so we want an answer very close to 4%, and answer C is the only plausible choice. Or we can calculate the probability exactly as in the post above.

Regor60 · Answer

Two digit numbers whose unit digits end with:

0,1,5 and 6

satisfy the conditions in the question.

There are 9 instances of each.

Number of ways to draw two numbers from each of the nine:

9!/2!7! = 36

Total number of ways:

4*36 = 144

Number of ways to draw two numbers from the 90 that span 10 to 99:

90!/2!88! = 45*89

Probability of drawing according to question:

144/(45*89) = 48/(15*89) =

16/(5*89) = 16/445

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Oppenheimer1945 · Answer

_1,_1 or _5,_5 or _6,_6 or _0,_0

4 pairs
10's digit =9*8 ways

Total ways=(9*8)*4

P=(9*8)*4/90*89
=32/890=16/445

joshmun2013 · Answer

To start, there are a total of 90 2-digit numbers.

Since it is without replacement, total cases would be 90*89 = 8010.

"two numbers have the same units digit and the product of the two numbers has the same units digit as the two numbers"

This condition is only met when the units digits are 0, 1, 5, and 6 (e.g. 1*1 = 1, 5*5 = 25, etc).

When the units digit is 1, there are 9 choices for the first number (e.g. 11, 21, 31, ... 91) and there are 8 choices left for the second number (excluding the number chosen for the first). So 8*9 = 72 ways.

Repeat for units digit 0, 5, 6. This gives a total of 72*4 = 288 ways.

288/8010 --> 144/4005 --> 16/445

Hence the answer is C.

Umber · Answer

Series of unit digits 1,2,5,6 from 1_ to 9_ it will be 9. Now 4 * 9C2 is our favorable outcomes divided by 90C2. Answer C

A super lotto has all the two digit numbers written on balls kept in

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