P( a student getting an A) = 5/20=1/4
There are 3 students. The probability that all three got an A is (1/4)(1/4)(1/4)=1/64.
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.
The probability that three F2 seeds chosen from Mendel's study group will have at least one yellow seed is 63/64. It would be very rare to get three green seeds.
hypergeometric distribution f(k;N,n,m) = f(3;52,4,3)
One would need more facts about the situation to be able to answer this question.
The probability of drawing the first ace is 4 in 52. The probability of getting the second ace is 3 in 51. The probability of getting the third ace is 2 in 50. The probability, then, of drawing three aces is (4 in 52) times (3 in 51) times (2 in 50), which is 24 in 132600, or 1 in 5525, or about 0.0001810
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.
The probability that three F2 seeds chosen from Mendel's study group will have at least one yellow seed is 63/64. It would be very rare to get three green seeds.
according to me its 1/2 as the possibility of getting an even is 1/2
It is 22800/59280 = 0.3846 approx.
hypergeometric distribution f(k;N,n,m) = f(3;52,4,3)
One would need more facts about the situation to be able to answer this question.
that 8 out of 13 chance.
50
4/52
The probability of drawing the first ace is 4 in 52. The probability of getting the second ace is 3 in 51. The probability of getting the third ace is 2 in 50. The probability, then, of drawing three aces is (4 in 52) times (3 in 51) times (2 in 50), which is 24 in 132600, or 1 in 5525, or about 0.0001810
About 18.5 percent thst both will have seatbelts on. (.43 x 2)
1 in 6