General Genetics PROBLEMS ON PROBABILITY 1. What are the probabilities that, fro
ID: 3067639 • Letter: G
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General Genetics PROBLEMS ON PROBABILITY 1. What are the probabilities that, from a normal pack of playing cards: A. A single card picked blind will be a diamond? B. A deuce of hearts will be removed followed by a diamond? C. A single card picked will be either a diamond or a spade? D. Four cards picked out will be two aces and two queens? (Assume that, in parts B and D, each card that is removed from the deck is replaced and the deck shuffled before the next card is removed.) 2. In man, the gene for freckles (F) is dominant to its allele (t) for absence of freckles. A certain couple, both heterozygous for this gene, have five children. What is the possibility that: A. None will have freckles? B. Three will have freckles and two will lack freckles? C. The first two will have freckles and the last three will lack freckles? D. All will be girls who have freckles? 3. A couple have two girls and are expecting their third child. They hope it will be a boy. What is the probability that their wish will be realized? 4. Another couple have eight children, all boys. What is the chance that their ninth child will be another boy? 5. In New York in 1935, of 148,462 deaths, 4,196 were of diabetes and 7,436 were of tuberculosis. Based on this data, what is the probability that the next person to die will die of diabetes? of diabetes or tuberculosis? 6. What is the probability that, of a family of five children: A. The first will be a girl? B. The first two will be girls and the last three will be boys? C. Two will be girls and three will be boys? 7. In tossing three coins simultaneously, what is the probability, in one tos of: A. Three heads. B. two heads and one tail? s, 8. Black hair in the guinea pig is dominant to white hair. In families of five offspring where both parents are heterozygous black, with what frequency would we expect to find: A. 3 whites and 2 blacks? B. 2 whites and 3 blacks? C. 1 white and 4 blacks? D. All whites?Explanation / Answer
Ans-4. Answer=1/2
Ans-5.
person die due to diabetes= 4196/148462 = 0.0282
Assuming no patient is common among Diabetes and tuberculosis
probability=p(diabetes)+p(tuberculosis)= (4196+7436) / 148462 = 0.0784
Ans-7
Sample space= {HHH,HHT,HTH,HTT,THH,THT,TTH,TTT}
p(three heads)= 1/8
p(two heads and one tail)= 3/8
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