It is (26/26)*(25/26)*(24/26)*(23/26) = 358800/459676 = 0.79 approx.
It is (5/26)4 = 625/456,976 = 0.00137 approx.
When an event is repeated, the probability of it occurring is squared. For instance, if an outcome had the probability of 1/4, then the outcome happening twice would have a probability of 1/16. Note, however, that this does not mean that the second event has different probabilities. That particular outcome will always be 1/4, regardless of anything that happened before it.
You have a total of 11 letters in "mathematics" and you have 4 vowels (a,e,a,i) so the probability of drawing a vowel is 4/11. In other words if you were to consider the vowel's to be 1's and the consonants 2's. What is the probability of drawing a "1". There would be 4 1's and 7 2's. It would be 4/11
Word 1) 'math' has one vowel letter among a total of 4 letters. The probability of randomly selecting the vowel letter 'a' is P(v) = 1/4. Word 2) 'jokes' has two vowel letters among a total of 5 letters. The probability of randomly selecting a vowel letter is P(v) = 2/5. The probability of randomly selecting a vowel letter from the first word and a vowel letter from the second word is: P(v1,v2) = 1/4 (2/5) = 2/20 = 1/10 = 0.10 = 10.0%
If it is a fair coin, the probability is 1/4.If it is a fair coin, the probability is 1/4.If it is a fair coin, the probability is 1/4.If it is a fair coin, the probability is 1/4.
It is (5/26)4 = 625/456,976 = 0.00137 approx.
When an event is repeated, the probability of it occurring is squared. For instance, if an outcome had the probability of 1/4, then the outcome happening twice would have a probability of 1/16. Note, however, that this does not mean that the second event has different probabilities. That particular outcome will always be 1/4, regardless of anything that happened before it.
If the numbers and letters can be repeated then there are 45,697,600 possible outcomes. If the letters and numbers can not be repeated there are 32,292,000 possible outcomes.
1/6
It is 4/7.It is 4/7.It is 4/7.It is 4/7.
Independent events, so P(both)=(5/26)(4/10)=0.07692307692. The 4/10 comes from the fact that 0, 3, 6, and 9 are the 4 digits that are multiples of 3.
4 because in word perseverance we have 4 letters E
To find the number of linear arrangements of the letters in "CALL," we need to consider the repetitions of letters. The word "CALL" has 4 letters where 'L' appears twice. The formula for arrangements of letters with repetitions is given by ( \frac{n!}{p1! \times p2! \times \ldots} ), where ( n ) is the total number of letters and ( p1, p2, \ldots ) are the frequencies of each repeated letter. Therefore, the number of arrangements is ( \frac{4!}{2!} = \frac{24}{2} = 12 ). Thus, there are 12 distinct arrangements of the letters in "CALL."
If the card is drawn at random, then the probability is 2/13.
The sequence appears to follow a pattern of repeated letters and increasing alphabetical order. After "ii," the next letters should continue the alphabetical sequence. The next four letters would be "jjkk."
To find the probability of randomly selecting the letters S, N, O, and T in that specific order from the letters in "FOUNDATIONS," we first note that there are 12 letters in total. The probability of selecting S first is 1/12, then N (1/11), O (1/10), and T (1/9). Therefore, the probability of selecting S, N, O, and T in that order is (1/12) * (1/11) * (1/10) * (1/9) = 1/11880, or approximately 0.000084.
If none of the letters were repeated, the answer would be 7!. However, the letters M & E are repeated. So, we need to divide the 7! by 2! twice to account for the two letters repeated twice. The solution is: 7! / (2! * 2!). This written out is: (7*6*5*4*3*2*1)/(2*1*2*1) or 1260.