91Ó°ÊÓ

It is given that:

Information regarding $187$is given in table.

Possible outcomes are $187$

In table, $104/187$eggs are assigned hot water.

$P\left(\mathrm{Hot}\right)=\frac{\text{Number of favourable outcomes}}{\text{Number of possible outcomes}}=\frac{104}{187}$

Also, in table $75/187$are assigned to hot water and hatched.

Hence, $P\left(\text{Hatched and Hot}\right)=\frac{\text{Numberof favourableoutcomes}}{\text{Number of possible outcomes}}=\frac{75}{187}$

Using conditional probability

$P(\text{Hatched}âˆ\pounds \text{Hot})=\frac{P\left(\text{HatchedandHot}\right)}{P\left(\text{Hot}\right)}=\frac{\frac{75}{187}}{\frac{104}{187}}=\frac{75}{104}≈0.7212=72.12\%$

Hence, $72.12\%$of eggs assigned to hot water hatched and probability is$0.7212$

We know that $P(\mathrm{B}âˆ\pounds \mathrm{A})=\frac{P(\mathrm{A}âˆ\copyright \mathrm{B})}{P(\mathrm{A})}=\frac{P(\mathrm{A}\text{and}\mathrm{B})}{P(\mathrm{A})}$

In table $129/187$eggs are hatched.

$⇒P\left(\text{Hatched}\right)=\frac{\text{Numberof favourable outcomes}}{\text{Number of possibleoutcomes}}=\frac{129}{187}$

Also in table, $16$hatched eggs are assigned to cold water and $38$hatched eggs are assigned to neutral water.

role="math" localid="1654797169661" $⇒54$hatched eggs are not assigned to hot water.

$P\left(\text{Hatched and not Hot}\right)=\frac{\text{Numberof favourableoutcomes}}{\text{Number of possible outcomes}}=\frac{54}{187}$

Using conditional probability:

$P\left(\text{Hatched|notHot}\right)=\frac{P\left(\text{Hatchedandnot Hot}\right)}{P\left(\text{Hatched}\right)}=\frac{\frac{54}{187}}{\frac{129}{187}}=\frac{54}{129}=\frac{18}{43}≈0.4186$

$=41.86\%$

About role="math" localid="1654797274671" $41.86\%$of the hatched eggs were not assigned to hot water and probability is$0.4186$