What is the probability that a fair coin lands Heads 4 times out of 5 flips? Out of bit strings of length ten, what is the probability that the bit string begins and ends with 0? What if the bit string has more 0s than 1s? and, what if the bit string has the sum of its digits equal to seven?
To find the probability of a specific outcome, we need to divide the number of favorable outcomes by the total number of possible outcomes.
1. Probability of getting heads 4 times out of 5 flips:
To calculate this, we can use the binomial probability formula. The probability of getting heads (H) on each flip is 1/2, and the total number of flips is 5. So, the probability of getting 4 heads is calculated as follows:
P(4 heads out of 5 flips) = (5 choose 4) * (1/2)^4 * (1/2)^(5-4)
The notation "n choose k" denotes the number of ways to choose k items from a set of n items. In this case, it is 5 choose 4 = 5! / (4! * (5-4)!) = 5.
Plugging in the values, we get:
P(4 heads out of 5 flips) = 5 * (1/2)^4 * (1/2)^1 = 5 * (1/16) * (1/2) = 5/32
Therefore, the probability of getting heads 4 times out of 5 flips is 5/32.
2. Probability of a bit string of length ten beginning and ending with 0:
In a bit string of length ten, the first and last bits are fixed, so we have 8 remaining positions to fill. Each position can be either a 0 or a 1, so there are 2^8 possible combinations for the remaining positions.
Since we want the first and last bits to be 0, only the remaining 8 positions matter. So, the probability that the bit string begins and ends with 0 is:
P(bit string begins and ends with 0) = 1/2 * 1/2 * (2^8)
Multiplying the probabilities, we get:
P(bit string begins and ends with 0) = 1/4 * 256 = 64.
Therefore, the probability that a bit string of length ten begins and ends with 0 is 64.
3. Probability of a bit string with more 0s than 1s:
For a bit string of length ten, the total number of possible combinations is 2^10. To find the number of bit strings with more 0s than 1s, we need to consider the different patterns.
If a bit string has more 0s than 1s, it can have 6, 7, 8, 9, or 10 zeros.
- For 6 zeros: We choose 6 positions out of 10, and the remaining positions can be filled with 1s. So, the number of bit strings with 6 zeros is (10 choose 6) = 10! / (6! * (10-6)!) = 210.
- Similarly, for 7, 8, 9, and 10 zeros, we get (10 choose 7) = 120, (10 choose 8) = 45, (10 choose 9) = 10, and (10 choose 10) = 1 respectively.
Adding up the number of bit strings with more 0s than 1s, we get:
Number of favorable outcomes = 210 + 120 + 45 + 10 + 1 = 386.
Therefore, the probability of a bit string having more 0s than 1s is:
P(more 0s than 1s) = 386 / (2^10) = 386 / 1024.
4. Probability of a bit string with the sum of its digits equal to seven:
To find the probability, we need to determine the number of bit strings with a sum of digits equal to seven.
Let's consider the possible combinations of 0s and 1s that add up to seven.
- For one 1 and six 0s: The 1 can appear in any of the seven positions, so we have (7 choose 1) = 7 possible combinations.
- Similarly, for two 1s and five 0s, we have (7 choose 2) = 21 possibilities.
Continuing this pattern, for three 1s, four 0s, we have (7 choose 3) = 35 combinations.
Adding up all these combinations, we get:
Number of favorable outcomes = 7 + 21 + 35 = 63.
Therefore, the probability of a bit string having the sum of its digits equal to seven is:
P(sum of digits = 7) = 63 / (2^7) = 63 / 128.