If the choice is unbiased, the change is 14/(10+14). If the chooser prefers choosing boys, the probability is 0.
10 :)
2/5
No heads means that every toss lands tails. (0.5)30= 9.3 x 10-10 Note that 109 = 1 trillion, so the probability can be stated this event is likely to occur about 9 times in 10 trillion tosses.
The ratio of girls to total students is 15:25, or 3:5. Three out of five students are girls so there would be a 60% probability that a girl would be chosen; a 2 out of 5 chance, or 40% probability that a boy would be chosen.
For each birth, you have two choices - either a boy or a girl. Then, the probability for a certain birth to obtain a choice is ½. Using Binomial Theorem, we have (10 choose 8)(½)8(½)² = 45/1024.
This is a Binomial Probability; p=0.5, n=10 & x=7. Since you want the probability of exactly 7, in the related link calculator, after placing in the above values, P(x=7) = 0.1172 or 11.72%.
If the choice is unbiased, the change is 14/(10+14). If the chooser prefers choosing boys, the probability is 0.
10 :)
This is one minus the probability of having no repeated letters. The probability of having no repeated numbers is 10/10 x 9/10 x 8/10 x 7/10 x 6/10 x 5/10 which equals 0.1512. The probability of having no repeated letters is 26/26 x 25/26 = 0.96153846153846153846153846153846 These multiplied together gives 0.14538461538461538461538461538462 1 minus this is 0.85461538461538461538461538461538 And so the probability of having a repeater letter or digit is roughly 0.855
2/5
17 out of 27, or about 0.6296.
Assuming that the probability of 1 woman have a boy is .5 Binomial Distribution can be used to solve the problem. b(k;n,p) = (n choose k) p^k * (1-p)^(n-k) k= number of boys n= number of trials p= probability of boy Assume p is equal to probability of having a boy First question we are looking for four boys. ie k=4 n = 10 =(10 choose 4) 0.5^4 * (1-.5)^4 = 0.8203 Question 2) At least 7 girls means 1 boy or 0 boys will be born Case 1) 0 boys born Use binomial (10 choose 0) 0.5^0 * (1-.5)^8 = 0.00390625 Case 2) 1 boy born (10 choose 1) 0.5^1 * (0.5)^7 =0.0390625 Case 1 + Case 2= 0.04296875 Probability of at least 7 girls
The probability of choosing 2 girls at random from group of 25 students of which10 are girls and 15 are boys is:P( 2 girls) = (10/25)∙(9/24) = 3/20 = 0.15 = 15%
More information is needed to determine the answer to the question. We need to know the probability of success or failure.
1/2
The probability is 15/25 = 3/5