# There are 15 students. 5 of them are boys and 10 of them are girls. If 5 students are chosen, what is the probability that there are at least 2 boys?

##### 2 Answers
Aug 13, 2016

Reqd. Prob.$= P \left(A\right) = \frac{567}{1001}$.

#### Explanation:

let $A$ be the event that, in the selection of $5$ students, at least $2$ Boys are there.

Then, this event $A$ can happen in the following $4$ mutually exclusive cases :=

Case (1) :

Exactly $2$ Boys out of $5$ and $3$ Girls ( =5students - 2 boys) out of $10$ are selected. This can be done in $\left({\text{_5C_2)(}}_{10} {C}_{3}\right) = \frac{5 \cdot 4}{1 \cdot 2} \cdot \frac{10 \cdot 9 \cdot 8}{1 \cdot 2 \cdot 3} = 1200$ ways.

Case (2) :=

Exactly $3 B$ out of $5 B$ & $2 G$ out of $10 G$.
No. of ways$= \left({\text{_5C_3)(}}_{10} {C}_{2}\right) = 10 \cdot 45 = 450$.

Case (3) :=

Exactly $4 B$ & $1 G$, no. of ways$= \left({\text{_5C_4)(}}_{10} {C}_{1}\right) = 50$.

Case (4) :=

Exactly $5 B$ & $0 G$ (no G), no. of ways$= \left({\text{_5C_5)(}}_{10} {C}_{0}\right) = 1$.

Therefore, total no. of outcomes favourable to the occurrence of the event $A = 1200 + 450 + 50 + 1 = 1701$.

Finally, $5$ students out of $15$ can be selected in ""_15C_5=(15*14*13*12*11)/(1*2*3*4*5)=3003 ways., which is the total no. of outcomes.

Hence, the Reqd. Prob.$= P \left(A\right) = \frac{1701}{3003} = \frac{567}{1001}$.

Enjoy Maths.!

Aug 13, 2016

Probability of at least 2 boys = P[(2 boys & 3 girls) + (3 boys & 2 girls) + (4 boys & 1 girl) + (5 boys & 0 girl)]$= 0.5663$

#### Explanation:

p_(2 boys &3 girls) = (C(5,2)xx(C(10,3)))/((C(15,5))
$= \frac{10 \times 120}{3003} = \frac{1200}{3003} = 0.3996$

p_(3 boys &2 girls) = (C(5,3)xx(C(10,2)))/((C(15,5))
$= \frac{10 \times 45}{3003} = \frac{450}{3003} = 0.1498$

p_(4 boys &1 girl) = (C(5,4)xx(C(10,1)))/((C(15,5))
$= \frac{5 \times 10}{3003} = \frac{50}{3003} = 0.0166$

p_(5 boys &0 girl) = (C(5,5)xx(C(10,0)))/((C(15,5))
$= \frac{1 \times 1}{3003} = \frac{1}{3003} = 0.0003$

Probability of at least 2 boys = P[(2 boys & 3 girls) + (3 boys & 2 girls) + (4 boys & 1 girl) + (5 boys & 0 girl)]

$= 0.3996 + 0.1498 + 0.0166 + 0.0003 = 0.5663$