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A six-digit is to be formed from the given numbers 1, 2, 3, 4, 5 and 6. Find the probability that the number is divisible by 4.
- a)3/17
- b)4/15
- c)4/19
- d)4/17
- e)None of these
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
A six-digit is to be formed from the given numbers 1, 2, 3, 4, 5 and 6...
For a number to be divisible by 4, the last two digit should be divisible by 4.
So possible cases – 12, 16, 24, 32, 36, 52, 56, 64 (last two digits)
So favourable outcomes = 24 +24 +24 +24 + 24+ 24+24+24 = 192
So p = 192/720 = 4/15
So possible cases – 12, 16, 24, 32, 36, 52, 56, 64 (last two digits)
So favourable outcomes = 24 +24 +24 +24 + 24+ 24+24+24 = 192
So p = 192/720 = 4/15
Most Upvoted Answer
A six-digit is to be formed from the given numbers 1, 2, 3, 4, 5 and 6...
2 nd 4 divided by 4 but 1,3,5,6 not divided by 4 so the sum of 1,3,5,6 is 15
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Community Answer
A six-digit is to be formed from the given numbers 1, 2, 3, 4, 5 and 6...
Solution:
To find the probability that the six-digit number is divisible by 4, we need to consider the following cases:
Case 1: Last two digits form a number divisible by 4.
There are three possibilities for the last two digits to form a number divisible by 4: 12, 24, and 56. For each of these possibilities, we can arrange the remaining four digits in 4! ways. Therefore, the total number of six-digit numbers divisible by 4 in this case is:
3 × 4! = 72
Case 2: Last two digits form a number not divisible by 4, but the third digit is even.
There are two possibilities for the last two digits to form a number not divisible by 4: 14 and 34. For each of these possibilities, we can choose the third digit from the set {2, 6}. We can then arrange the remaining three digits in 3! ways. Therefore, the total number of six-digit numbers divisible by 4 in this case is:
2 × 2 × 3! = 24
Total number of six-digit numbers that can be formed from the given set of numbers:
6! = 720
Therefore, the probability that a six-digit number formed from the given set of numbers is divisible by 4 is:
(72 + 24)/720 = 4/15
Hence, the correct answer is option (b) 4/15.
To find the probability that the six-digit number is divisible by 4, we need to consider the following cases:
Case 1: Last two digits form a number divisible by 4.
There are three possibilities for the last two digits to form a number divisible by 4: 12, 24, and 56. For each of these possibilities, we can arrange the remaining four digits in 4! ways. Therefore, the total number of six-digit numbers divisible by 4 in this case is:
3 × 4! = 72
Case 2: Last two digits form a number not divisible by 4, but the third digit is even.
There are two possibilities for the last two digits to form a number not divisible by 4: 14 and 34. For each of these possibilities, we can choose the third digit from the set {2, 6}. We can then arrange the remaining three digits in 3! ways. Therefore, the total number of six-digit numbers divisible by 4 in this case is:
2 × 2 × 3! = 24
Total number of six-digit numbers that can be formed from the given set of numbers:
6! = 720
Therefore, the probability that a six-digit number formed from the given set of numbers is divisible by 4 is:
(72 + 24)/720 = 4/15
Hence, the correct answer is option (b) 4/15.
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