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Electric charges and coulomb's law (Basic)
10 Questions 10 Marks 10 Mins
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\(\left(\frac12\right)^2\)\(\left(1-\frac12\right)^{6-2}\)
= \(\frac{15}{64}\).
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Three flips of a fair coin
Suppose you have a fair coin: this means it has a 50% chance of landing heads up and a 50% chance of landing tails up.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.
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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 ...
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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 ...
The probability of getting heads on the toss of a coin is 0.5. If we consider all possible outcomes of the toss of two coins as shown, there is only one outcome ...
Probability of an event, P(E) = number of favourable outcomestotal number of outcomes. ⇒ P(E) = 12. Therefore, the probability of getting a head is 12 .
답변 1개 · 인기 답변: The correct option is D 1/2 Given, Sample space = Total number of outcomes = SH,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 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 ...
So if you flip a coin 10 times in a row-- a fair coin-- you're probability of getting at least 1 heads in that 10 flips is pretty high. It's 1,023 over 1,024.
Khan Academy · 2013. 1. 5.
In this task you can use the rule called Bernoulli's Scheme.
In this scheme you repeat an experiment which can end with one of 2 results (usually called a success and a failure) and want to calculate the probability of getting exactly #k# "success" results.
The probability can be calculated as:
#P(S_k)=((n),(k))p^k(1-p)^(n-k)#, where:
#n# is a total number of experiments
#k# is an
expected number of successes
#p# is the probability of a success
The first task can be simply solved by calculating the probability of 2 successes:
A "success" is "tossing heads in a single toss"
A "failure" is "tossing tails in a single toss"
The probability of a success is #1/2#
The number of all experiments is #n=6#
The number of successes is #k=2#
#P(S_2)=((6),(2))(1/2)^2(1/2)^(4)=((6),(2))*(1/2)^6=(6!)/(4!*2!)(1/64)=15/64#
In the second part you have to calculate #P(S_2)+P(S_1)+P(S_0)#
#P(S_0)+P(S_1)+P(S_2)=#
#((6),(0))(1/2)^6+((6),(1))(1/2)^6+((6),(2))(1/2)^6=#
#1*1/64+6*1/64+15*1/64=22/64=11/32#