Thus, the entire shape is made of 9 identical triangles, 3 of them shaded. Therefore, the chance of hitting them in a random throw of a ball is 3/9 = 1/3.
Since we want a "5" AND a "2", multiply the probabilities to get 1/6 × 1/6 = 1/36. The same goes for all other scenarios - each of them has a probability of 1/36.
For "At least questions, it is easier to calculate the odds of NOT HITTING the target on both shots, and then subtracting the probability of this happening from 1: The probability of NOT hitting twice is equal to the probability of not Hitting on the first throw AND not hitting on the second throw.
Probability of one means something MUST happen. Probability value is always between zero and one, inclusive. In other words, it is a fraction, and as such may be also expressed as a decimal or percent. In the GMAT, probability problems are just another way for GMAC to test part-to-whole relationships.
In the GMAT, probability problems are just another way for GMAC to test part-to-whole relationships. You are likely to see 1-2 probability problems in your GMAT.
This means it is never going to happen - a probability of zero. Probability of zero means something can never happen. Probability of one means something MUST happen. Probability value is always between zero and one, inclusive. In other words, it is a fraction, and as such may be also expressed as a decimal or percent.
There is one wanted outcome (getting the number "2") out of 6 possible outcomes (the numbers 1-6).
Probability means the chance or likelihood that something may happen. For instance: What is the probability of getting a result of "1" when rolling a regular six-sided die?
According to Stat. (1), 80% of the students do not get housing in any given year, thus 20% do get housing. The probability of winning the housing lottery in a certain year is thus 1/5 (=20%). You're asked to find the probability that the student will receive housing in AT LEAST one of the two years.
The effective range is actually 8 to 96 inclusive.
An outcome is one thing that may happen in the setup of the problem.
The average score of 100 students taking a statistics final was 70, with a standard deviation of 7. Assuming a normal distribution, what test score separates the top 5% of the students from the lower 95% of the students?
A binomial distribution has 50 trials (n = 50) with a probability of success of 0.50 (π = 0.50). We would like to find the probability of 34 or more successes using the normal distribution to approximate the binomial. Applying the continuity correction factor, what z-score should be used?
based on convenience sample, a researcher used SPSS and calculated a mean age of 50 with a standard error of 3. Thus, the 68% confidence interval is about 47 to 53
It is 1 - the probability of getting no heads.
To calculate the probability of getting at least one head you can either calculate the probability of getting 1 head, 2 heads & 3 heads and add them up to get the required probabilty. The other method is to calculate the probability of getting no head (the complement of the required event) and them subtract this probabilty from 1 to get the probability of getting at least one head.
The sum of the probability of the occurence of an event and the probability of the occurence of its complement is 1. This is the complement rule. You can understand this by this image:
When you toss 3 coins there are 8 equally likely outcomes. This is found by taking the number of outcomes, 2; heads or tails, and raising it to the power of the number of events, three. 2^3 = 8. To get exactly one head there are three locations to place the one head, either 1st, 2nd or 3rd.
Three coins give 2^3 = 8 outcomes. With fair coins, each has a probability of 1/8.