Probability maths question . need asap

a) 1/22

b) 3/44

c) 3/11.

Step-by-step explanation:

a).

Prob(picking a blue first) = 5/12.

Prob(picking a yellow next) = 4/11    ( as it is without replacement)

Prob(purple next) = 3/10

Probability of picking these in this order = 5/12 * 4/11 * 3/10

= 1/22 (answer).

Note the probabilities are multiplied because the 3 events are independent.

b)

Prob(all the same colour) = Prob(All are blue) + Prob(all are yellow) + Prob ( All are purple)

Prob(All are blue) = 5/12 * 4/11 * 3/10 =  1/22

Prob(all are yellow) = 4/12 * 3/11 * 2/10 = 1/55

Prob(all purple) = 3/12 * 2/11 * 1 /10 = 1/220

So probability there are all the same colour = the sum of the above

= 3/44 (answer).

c)  I take this to mean that all 3 are a different colour.

This will be the number of combinations of blue, yellow and purple possible which is 3! = 6.

So the answer is 6 * 1/22 = 3/11.

answer: the answer is one

step-by-step explanation:

pemdas, in this situation there is no parenthesis or exponents so you go to multiplication, 1x1=1, then division 1/1= 1 then addition 1+1=2 then subtraction 2-1=1

an = 22 + (n-1) *6

110 seats

step-by-step explanation:

an=a1+(n-1)(d)

a1 = 22   this is the first item in the sequence

d =6   this is how the sequence goes up

a)   an = 22 + (n-1) *6

b) a12 = 22 + (12-1) *6

    a12 = 22+11*6

  a12 = 22 +88

  a12 = 110