Will mark the Brainliest !!!!! A team of 5 boys and 4 girls will be chosen from a group of 16 boys and 13 girls,
1. If Bob is one of the students, how likely is he to be picked?
2. If Jane is also a student, how likely is it that Jane and Bob will both be picked?
3. How likely is it that at least Jane or Bob will be picked?
4. How many different teams are possible?

Bob will be one of the fixed boys to be picked. Hence, actually 4 boys are to be picked from 15. The permutations of girls being picked remains the same.

Probability = Permutations with Bob as one of the boys / Total permutations

Probability = (¹⁵P₄*¹³P₄) / (¹⁶P₅*¹³P₄) = ¹⁵P₄ / ¹⁶P₅

Probability = [tex]\frac{15!}{(15-4)!}/\frac{16!}{(16-5)!}[/tex] = 15! / 16! = 1/16

2.

Now, Bob is one of the fixed boys and Jane is one of the fixed girls. Hence, actually 4 boys are to be picked from 15 and 3 girls are to be picked from 12.

Probability = Permutations with bob as one of the boys and jane as one of the girls / Total permutations

Probability = (¹⁵P₄*¹²P₃) / (¹⁶P₅*¹³P₄) = (1/16)*(1/13) = 1/208

3.

Now, the probability that at least Jane or Bob will be picked has been asked. This probability is a combination of three probabilities:

Probability = (Probability that only Bob will be picked) + (Probability that only Jane will be picked) + (Probability that both will be picked)

Probability = 1/16 + 1/13 + 1/208 = 0.123

4.

Total teams possible = ¹⁶P₅*¹³P₄ = 8994585600 teams are possible