Question

Will mark the Brainliest !!!!! A team of 5 boys and 4 girls will be chosen from a group of 16 boys and 13 girls,

Answer 1

Answer:

Step-by-step explanation:

The total number of permutations of boys and girls on the team are:

¹⁶P₅*¹³P₄

1.

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 = $\frac{15!}{(15-4)!}/\frac{16!}{(16-5)!}$ = 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