91影视

Given data,

To find the probability that the specimen contains type O blood or comes from the Hawaiian-Chinese ethnic group, that

$P(O鈭�HC)=P\left(O\right)+P\left(HC\right)-P(O鈭�HC)$

$=\frac{62337}{145057}+\frac{5385}{145057}-\frac{2206}{145057}$

$=0.4297+0.03712-0.0152$

$=0.4516$

Given data:

To determine the probability that the samples include type blood AB, assuming that it is from the Hawaiian ethnic group,

$P(AB鈭�H)=\frac{P(H鈭�AB)}{P\left(H\right)}$

$=\frac{99}{1903+2490+178+99}$

$=0.0212$

Given data,

The probability of having the blood group $B$is therefore,

$P\left(B\right)=\frac{17604}{145057}$

$=0.1214$

The probability of Hawaiians having blood group B is as follows:

$P(B鈭�\text{Hawaiian})=\frac{178}{4670}$

$=0.0381$

If the two events are independent, their probabilities should be equal. Because the two probability aren't equal, the occurrences "having blood type B" and "Hawaiian ethnic group" aren't mutually exclusive.

Given data:

The chance of not obtaining an A from $W$is the complement of getting an A from $W$on the first draw, with a total population of $145057$.

$1-\frac{50008}{145057}$

$=0.6553$

Given that A did not come from W in the initial draw, there are $145057$persons left, with $50008$in the A and W group. Because the population was so huge, these two odds are nearly equal.

$1-\frac{50008}{145057}$

localid="1652866656927" $=0.6553$

Use the multiplication rule, as follows, to get the likelihood of both events happening.

localid="1653040234665" $P\left(\text{None}\right)=0.{6553}^2$

$=0.4294$

We have,

$P(鈮�1)=1-P\left(\text{None}\right)$

localid="1652866823781" $=1-0.429$

$=0.5706$