2022. 1. 17. — Probability of getting one head = 1/2. Here, tossing a coin is an independent event, its not dependent on how many times it has been tossed. Show 답변 1개 A coin has 2 faces: Head (H) and Tail (T). So total number of outcomes are 2 H or T And we are interested to know the probability of occuring of head (So ... 2020. 9. 30.답변 1개 The coins are “fair” so there's an equal probability of a coin producing either a head or a tail when tossed. So each coin has a probability of 0.5 (or 50% ... The probability of getting heads on the toss of a coin is 0.5. Hope You Got it !✔︎. active attachment. 답변 10개 · 5표: Answer: The probability of getting heads on the toss of a coin is 0.5. If we consider all ... 답변 1개 The correct option is D. 12. Given,. Sample space = Total number of outcomes = S{H,T} Number of favourable outcomes = 1. Probability of an event, P(E) = number ... 2019. 11. 17. — Every toss is independent, so the probability of a fair coin landing on heads will be always 0.5. Note that it is perfectly normal to get ... 답변 1개 · 1표: If I understand your question correctly you are asking whether the 4th toss is dependent on the first 3 - the answer is no. Every toss is independent, ... In counting the number of heads in 4 coin flips, the probability that we get exactly one head is the probability that we get anyone of the following 4 outcomes: ... since the coin is fair and the flips are independent. In fact, all 8 probabilities work out the same way. We always get 1/8. In other words, each of the 8 ... In our case, n = 5, p = 1 / 2, and k = 1. The probability of getting head exactly 5 times in 8 throws is 8C 5( 21) 8= 2 856= 327. Why the probability is 1/2 ... Correct Answer - Option 3 : \(\frac{15}{64}\) Concept: Binomial Distribution: If ‘n’ and ‘p’ are the parameters, then ‘n’ denotes the total number of times the experiment is conducted and ‘p’ denotes the probability of the happening of the event. The probability of getting exactly ‘k’ successes in ‘n’ independent trials for a Random Variable X is expressed as P(X = k) and is given by the formula: P(X = k) = nCk pk (1 - p)n - k Calculation: The probability of getting a head in a single toss is p = \(\frac12\). ∴ The probability of getting 2 heads in 6 tosses, will be: P(X = 2) = 6C2\(\left(\frac12\right)^2\)\(\left(1-\frac12\right)^{6-2}\) = \(\frac{15}{64}\). When in doubt about an exercise, a reasonable mathematical approach is to simplify the exercise data if possible; here, let's look for the probability of getting exactly $2$ heads when a coin is tossed $a=3$ times. Let's describe this random experience with the sample space $S=\{HHH,HHT,HTH,HTT,THH,THT,TTH,TTT\}$. By the way, we make sure that $Card(S)=2^a=2^3=8$. The event $A':$"getting exactly $2$ heads" is $A'=\{HHT,HTH,THH\}$. We have$Card(A')=\mathrm{C}_{3}^{2}=\mathrm{C}_{3}^{1}=3$. Suppose that the outcomes in the sample space are equally likely to occur. Then $p(A')=\frac{\mathrm{C}_{3}^{2}}{2^3}=\frac{\mathrm{C}_{3}^{1}}{2^3}$. So back to our original exercise, and based on what we have learned in our simple case, $p(A)=\frac{\mathrm{C}_{6}^{4}}{2^6}=\frac{\mathrm{C}_{6}^{2}}{2^6}=\frac{\frac{6\times5}{2!}}{2^6}$. Want to join the conversation?
What is the probability of getting exactly 2 heads in 6 tosses of a fair coin?So we have a-- what is this-- a 3/8 chance of getting exactly two heads.
What is the probability of getting exactly 3 heads in 6 tosses of a fair coin?Probability of getting 3 heads : P(X=3)=6C(21)3(21)6−3=20×21=165.
What is the probability of getting exactly 2 heads?The probability of getting two heads on two coin tosses is 0.5 x 0.5 or 0.25.
What is the probability of tossing a coin 6 times?There is a probability of 0.5 for each tail, and since each coin toss is independent, we can multiply that by itself six times in order to find the probability that each toss results in tails. Therefore, the probability of at least one head is as follows. So, there is a 98.44% probability of getting at least one head.
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