## Repeated Trials in probabilityRepeated trials refer to the outcomes of events that repeat one or more times. The probability of success (p) and failure (q) are used to determine the total probability of n successes in a trial. For example, The probability of tossing a coin thrice We know that a coin has two possible outcomes, Head (H) and Tail (T). The probability of the occurrence of Head is P(H) =1/2. The probability of the occurrence of Tail is P(T) = 1/2. The probability of Head (H) and Tail (T) outcomes is equal because both have equal chances to occur. The outcomes of tossing a coin thrice are (H, H, H), (H, H, T), (H, T, H), (T, H, H), (T, T, H), (T, H, T), (H, T, T), and (T, T, T). The outcomes of getting two heads and one tail = (H, H, T), (H, T, H), (T, H, H) Probability of each outcome will be (1/2 x 1/2 x 1/2) = (1/2) Total probability of getting two heads and one tail = The outcomes of tossing a coin twice are (H, H), (H, T), (T, H), (T, T). Similarly, we can find the probability of various outcomes. If a trial is repeated n time, the probability of 'r' successes and failures 'n - r' will be:
Where, p = probability of success q = probability of failure An experiment that repeats identically is known as ## Binomial DistributionIt is concerned with the repeated trials where the
Where, p = probability of success q = probability of failure n is the total number of trials r refers to the success in series of n trials. The value of r lies between 0 and n. The probability of 0 success is represented as: The probability of 1 success is represented as: The probability of 2 successes is represented as: The probability of n success is represented as: ( The sum of probabilities is always 1
= q
## ExamplesLet's discuss some numerical examples based on the repeated trials.
Probability that a pencil is defective = 2/10 = 0.2 Probability that a pencil is not defective = (1 - 0.2) = 0.8 Here, p = 0.2 q = 0.8 n = 10 a) Probability that exactly three will be defective is: = = b) Probability that atleast two will be defective is: = 1 - [(probability that none is defective) + (probability that one is defective)] = 1 - ( = c) Probability that none will be defective is: = = =
The number of success trials of getting heads = 5 Probability of success = ½ = 0.5 Probability of failure = (1 - 0.5) = 0.5 Using the binomial distribution, the probability of 'r' successes in series of n trials is given by:
where, n = 15 r = 6 p = 0.5 q = 0.5 Substituting the values, we get:
After solving, we get: =
P (Head) = 1/2 P (Tail) = 1/2 By Binomial distribution, the probability of 10 heads and 6 tails in total 16 tosses of a coin will be: P (X = 10) = P (X = 10) = P (X = 10) = P (X = 10) = P (X = 10) = 0.122 The expected number of cases in 256 sets is: 256 x P (X = 10) = 256 x 0.122 = 31.28 Thus, the cases one can expect 10 heads and 6 tails in 256 sets of 16 tosses of a coin are
P (Head) = 1/2 P (Tail) =1/2 By Binomial distribution, the probability of 4 heads and 2 tails in total 6 tosses of a coin will be: P (X = 4) = P (X = 4) = P = 0.234 The expected number of cases in 16 sets is: 16 x P (X = 4) = 16 x 0.234 = 3.75 Thus, the cases one can expect 4 heads and 2 tails in 16 sets of 6 tosses of a coin are
Mean = np Where, n is the number of items p is the probability that a sample is defective np = 2 20p = 2 P = 2/20 = 0.1 The probability that a sample is defective = 0.1 The probability that a sample is not defective = (1 - 0.1) = 0.9 The probability of atleast two defective in a sample of 20 is: 1 - (Probability that none is defective + probability that one is defective) = 1 - ( = 1 - ((0.9) = 1 - (0.3917) = 0.608 Thus, number of samples having atleast two defective out of 2000 samples is: 2000 x 0.608 = |

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