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If 4 whole numbers are taken at random and multiplied together, what is the chance that the last digit in the product is 1, 3, 7 or 9?
- a)15/653
- b)12/542
- c)16/625
- d)17/625
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
If 4 whole numbers are taken at random and multiplied together, what i...
3,7,9 are not possible in the unit place wen 4 whole numbers are multiplied togethr,only 1 i s possible,if the number chosen hav 1,3(3*3*3*3=81-1 as unit place),7(7*7*7*7=1 in unit place),9(9*9*9*9=1 in unit place)
so,possibilities of 1,3,7,9 in unit place of 4 number is 4C1*4C1*4C1*4C1=4*4*4*4=256..
thus numbers are=>1,11,21,31,41,51,61,71,81,91 with 1 in unit place.:=10 possibilities.
similarly,3,13,23,33,43,53,63,73,83,93:=10 possibilities
likewise 7,17,27,37,47,57,67,77,87,97:=10 poss.
and 9,19,29,39,49,59,69,79,89,99:=10 poss.
total posibilties:=10*10*10*10=10000
probability is 4^4/10^4=256/10000=16/625
Most Upvoted Answer
If 4 whole numbers are taken at random and multiplied together, what i...
Probability of getting a number with last digit as 1, 3, 7 or 9
To get a number with last digit as 1, 3, 7 or 9, we need to multiply two such numbers. The probability of getting a number with last digit as 1, 3, 7 or 9 is 4/10 or 2/5. So, the probability of getting two such numbers is (2/5) x (2/5) = 4/25.
Total number of combinations
There are a total of 4 whole numbers taken at random. So, the total number of combinations is 4C4 = 1.
Total number of combinations with last digit as 1, 3, 7 or 9
To get the total number of combinations with last digit as 1, 3, 7 or 9, we need to count the number of ways we can choose two numbers with last digit as 1, 3, 7 or 9 and two numbers with any other last digit. There are 2 numbers with last digit as 1, 3, 7 or 9 and 8 numbers with any other last digit. So, the total number of such combinations is 2C2 x 8C2 = 28.
Probability of getting a product with last digit as 1, 3, 7 or 9
The probability of getting a product with last digit as 1, 3, 7 or 9 is the ratio of the total number of combinations with last digit as 1, 3, 7 or 9 to the total number of combinations. So, the probability is 28/1 = 28.
Simplifying the probability
The probability can be simplified by reducing the numerator and denominator by 4. So, the probability is (2/25) / (1/4) = 8/100 = 2/25.
Final answer
The probability of getting a product with last digit as 1, 3, 7 or 9 is 2/25 or 16/625. Therefore, the correct answer is option 'C'.
To get a number with last digit as 1, 3, 7 or 9, we need to multiply two such numbers. The probability of getting a number with last digit as 1, 3, 7 or 9 is 4/10 or 2/5. So, the probability of getting two such numbers is (2/5) x (2/5) = 4/25.
Total number of combinations
There are a total of 4 whole numbers taken at random. So, the total number of combinations is 4C4 = 1.
Total number of combinations with last digit as 1, 3, 7 or 9
To get the total number of combinations with last digit as 1, 3, 7 or 9, we need to count the number of ways we can choose two numbers with last digit as 1, 3, 7 or 9 and two numbers with any other last digit. There are 2 numbers with last digit as 1, 3, 7 or 9 and 8 numbers with any other last digit. So, the total number of such combinations is 2C2 x 8C2 = 28.
Probability of getting a product with last digit as 1, 3, 7 or 9
The probability of getting a product with last digit as 1, 3, 7 or 9 is the ratio of the total number of combinations with last digit as 1, 3, 7 or 9 to the total number of combinations. So, the probability is 28/1 = 28.
Simplifying the probability
The probability can be simplified by reducing the numerator and denominator by 4. So, the probability is (2/25) / (1/4) = 8/100 = 2/25.
Final answer
The probability of getting a product with last digit as 1, 3, 7 or 9 is 2/25 or 16/625. Therefore, the correct answer is option 'C'.
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If 4 whole numbers are taken at random and multiplied together, what is the chance that the last digit in the product is 1, 3, 7 or 9?a)15/653b)12/542c)16/625d)17/625Correct answer is option 'C'. Can you explain this answer?
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