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It is given that:

Assuming:

A: About right

O: Overweight

U: Underweight

F: Female

As per Conditional Probability:

$P(\mathrm{B}鈭�\mathrm{A})=\frac{P(\mathrm{A}鈭�\mathrm{B})}{P(\mathrm{A})}=\frac{P(\mathrm{A}\text{and}\mathrm{B})}{P(\mathrm{A})}$

Information for $1200$college students is given.

Possible outcomes are $1200$

From table, $85/1200$college students thinks that their body image is about right.

$鈬�P(\mathrm{A})=\frac{\text{Number of favourable outcomes}}{\text{Number of possible outcomes}}=\frac{855}{1200}$

Also, $560/1200$female students think that their body image is about right.

$鈬�P(\mathrm{A}\text{and}\mathrm{F})=\frac{\text{Number of favourableoutcomes}}{\text{Number of possible outcomes}}=\frac{560}{1200}$

Using Conditional Probability:

$P(\mathrm{A}鈭�\mathrm{F})=\frac{P\left(\text{A and}\mathrm{F}\right)}{P(\mathrm{A})}=\frac{\frac{560}{1200}}{\frac{855}{1200}}=\frac{560}{855}=\frac{112}{171}鈮�0.6550=65.50\%$

It implies that $65.50\%$of students are of the view that their body image as about right are female students and the probability is$0.6550$

As per complement rule,

$P\left({\mathrm{A}}^{\mathrm{c}}\right)=P(not\mathrm{A})=1-P(\mathrm{A})$

According to conditional probability

$P(\mathrm{B}鈭�\mathrm{A})=\frac{P(\mathrm{A}鈭�\mathrm{B})}{P(\mathrm{A})}=\frac{P(\mathrm{A}\text{and}\mathrm{B})}{P(\mathrm{A})}$

In table, $760/1200$college students are females.

$鈬�P(\mathrm{F})=\frac{\text{Number of favourableoutcomes}}{\text{Number of possibleoutcomes}}=\frac{760}{1200}$

Also, $163/1200$female students think that their body weight is about overweight.

$鈬�P\left(\mathrm{O}\text{and}\mathrm{F}\right)=\frac{\text{Number of favourableoutcomes}}{\text{Number of possible outcomes}}=\frac{163}{1200}$

Using Conditional Probability

$P(\mathrm{O}鈭�\mathrm{F})=\frac{P\left(\mathrm{O}\text{and}\mathrm{F}\right)}{P(\mathrm{F})}=\frac{\frac{163}{1200}}{\frac{770}{1200}}=\frac{163}{760}$

Using Complement Rule $P\left((\mathrm{O}鈭�\mathrm{F}{)}^{\mathrm{c}}\right)=1-P(\mathrm{O}鈭�\mathrm{F})$

$=1-\frac{163}{760}$

$=\frac{597}{760}鈮�0.7855=78.55\%$

Hence, $78.55\%$do not perceive their body image as overweight are female students and the probability is$0.7855$