How do I solve these probabilities problems?
An experiment consists of placing the numbers 1-50 in a hat. If 2 numbers are drawn, find the probability of drawing:
1. a number with 2 digits that are the same followed by a single-digit number with replacement.
2. a number divisible by 10 followed by a number divisible by 5 without replacement.
3. a one-digit number followed by a number divisible by 3 in consecutive draws without replacement.
An experiment consists of placing the numbers 1-50 in a hat. If 2 numbers are drawn, find the probability of drawing:
1. a number with 2 digits that are the same followed by a single-digit number with replacement.
2. a number divisible by 10 followed by a number divisible by 5 without replacement.
3. a one-digit number followed by a number divisible by 3 in consecutive draws without replacement.
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1. Out of the 50 numbers, 4 numbers have two digits that are the same, and 9 numbers are single-digit numbers.
P(# with two digits the same, then single digit #, w/ replacement)
= (4/50)(9/50)
= 9/625.
2. Out of the 50 numbers, 5 numbers are divisible by 10, 10 numbers are divisible by 5, and all numbers divisible by 10 are divisible by 5. So selecting a number divisible by 10 on the first draw reduces, by one, the number of remaining numbers divisible by 5 for the second draw.
P(# divisible by 10, followed by # divisible by 5, w/o replacement)
= (5/50)(9/49)
= 9/490.
3. Out of the 50 numbers, 9 numbers are one-digit, 16 numbers are divisible by 3, 3 of the one-digit numbers are divisible by 3, and 6 of the one-digit numbers are not divisible by 3.
Selecting one of the 3 one-digit numbers divisible by 3 on the first draw reduces, by one, the number of remaining numbers divisible by 3 for the second draw.
Selecting one of the 6 one-digit numbers not divisible by 3 on the first draw does not affect the number of remaining numbers divisible by 3 for the second draw.
P(one-digit #, followed by # divisible by 3, w/o replacement)
= (3/50)(15/49) + (6/50)(16/49)
= 9/490 + 48/1225
= 141/2450.
Lord bless you today!
P(# with two digits the same, then single digit #, w/ replacement)
= (4/50)(9/50)
= 9/625.
2. Out of the 50 numbers, 5 numbers are divisible by 10, 10 numbers are divisible by 5, and all numbers divisible by 10 are divisible by 5. So selecting a number divisible by 10 on the first draw reduces, by one, the number of remaining numbers divisible by 5 for the second draw.
P(# divisible by 10, followed by # divisible by 5, w/o replacement)
= (5/50)(9/49)
= 9/490.
3. Out of the 50 numbers, 9 numbers are one-digit, 16 numbers are divisible by 3, 3 of the one-digit numbers are divisible by 3, and 6 of the one-digit numbers are not divisible by 3.
Selecting one of the 3 one-digit numbers divisible by 3 on the first draw reduces, by one, the number of remaining numbers divisible by 3 for the second draw.
Selecting one of the 6 one-digit numbers not divisible by 3 on the first draw does not affect the number of remaining numbers divisible by 3 for the second draw.
P(one-digit #, followed by # divisible by 3, w/o replacement)
= (3/50)(15/49) + (6/50)(16/49)
= 9/490 + 48/1225
= 141/2450.
Lord bless you today!