Question

There is a bag filled with 3 blue and 4 red marbles.

Answer 1

Answer:

The probability of getting at least 1 blue is 5/7

Step-by-step explanation:

Here we have the probability that the first is blue and the second is red is thus;

p(First Red Second Red) = 4/7 × 3/6 = 2/7

Therefore, the probability of getting at least 1 blue, is the complement of getting both red

That is, P(At least 1 blue) = P'(both red) = 1 - P(Both red) = 1 - 2/7 = 5/7

The question can also be answered by solving in steps as follows;

P(First blue, second blue) = p(first blue) × p(second blue $|$  first blue) = 3/7×2/6 = 1/7

P(First blue, second red) = p(first blue) × p(second red $|$  first blue) = 3/7×4/6 = 2/7

P(First red, second blue) = p(first red) × p(second blue $|$  first red) = 4/7×3/6 = 2/7

Total probability = P(at least 1 blue) = P(First blue, second blue) + P(First blue, second red) + P(First red, second blue)

∴ Total probability = P(at least 1 blue) = 1/7 + 2/7 + 2/7 = 5/7.

Answer 2

Answer:

Pr (at least 1 blue ball) = 4/7

Step-by-step explanation:

Number of blue balls= 4

Number of red balls= 3

Total number of balls = 3+4 = 7

This is a case of probability without replacement since the ball taken at random isn't replaced.

Pr (blue ball) = 4/7

Pr (red ball) = 3/7

Where Pr means probability

Pr (at least 1 blue ball) = Pr (1 blue ball and 1 red ball) + Pr (2 blue balls)

In probability without replacement, after picking the first ball, only the total number of balls would change if the colours are different. It would reduce by 1.

But If the balls are the same, the total number of balls would reduce by 1 and the total number of same ball would also reduce by 1

Pr (1 blue ball and 1 red ball) = (4/7 × 3/6)

Pr (2 blue ball) = (4/7 × 3/6)

Pr (at least 1 blue ball) = (4/7 × 3/6) + (4/7 × 3/6)

Pr (at least 1 blue ball) = 12/42 + 12/42

Pr (at least 1 blue ball) = 24/42 = 4/7

Pr (at least 1 blue ball) = 4/7