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# counting problem

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Sam writes down the numbers \$1,\$ \$2,\$ \$\dots,\$ \$315,\$ \$316,\$ \$317,\$ \$\dots,\$ \$248,\$ \$249,\$ \$250.\$

(a) How many digits did Sam write, in total?

(b) Sam chooses one of the digits written down at random.  What is the probability that Sam chooses a \$2\$?

Mar 19, 2024

#1
+624
+1

Absolutely, I've been improving my problem-solving abilities in counting problems. Let's tackle this step-by-step.

(a)  (Observe the digits of the numbers 1 to 250).

Numbers from 1 to 9 have 1 digit each (9 numbers).

Numbers from 10 to 99 have 2 digits each (90 numbers).

Numbers from 100 to 250 have 3 digits each (151 numbers).

Therefore, the total number of digits is 9×1+90×2+151×3=9+180+453=642.

(b) (Calculate the number of times the digit 2 appears)

Each number from 10 to 19 contains the digit 2 once. (10 numbers)

Each number from 20 to 29 contains the digit 2 once. (10 numbers)

Similarly, each hundred from 100 to 240 contains the digit 2 ten times (15 hundreds * 10 = 150 times)

Additionally, numbers 200, 210, 220, 230, and 240 each contain the digit 2 twice. (5 numbers * 2 = 10 times)

So, the total number of times the digit 2 appears is 10+10+150+10=180.

(c)  The probability of Sam choosing a 2 is the number of times 2 appears divided by the total number of digits:

Probability = (Number of 2s) / (Total number of digits) = 180 / 642 = 20/71

Therefore, the probability of Sam choosing a 2 is 20/71.

Mar 19, 2024
#2
+128794
+1

No. of 2's

In the ones place

10 in the first hundred

10 in the second hundred

5 in last 50

= 25

In the tens place

10 in the first hundred

10 in the second hundred

10 in the last 50

= 30

In the hundreds place

51  in  the last 51

Total 2's  = [ 25 + 30 + 51  ]  =  106

Prob of a 2  =  106  /  642  =    53  /  321

Mar 19, 2024