Question

A box contains 3 blue marbles, 4 red, 6 green marbles and 2 yellow marbles. If two marbles are picked at random, what is the probability that they are either blue or yellow ?

Answer 1

Answer: 5/15 = 1/3

Step-by-step explanation:

no. of blue marbles=3

no. of yellow marbles=2

P(Blue or Yellow)= 3+2/15

                           =5/15 =1/3

Answer 2

Hi there!
Here's the answer:

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In a Box,
No. of Blue marbles = 3
No. of Red Marbles = 4
No. of green Marbles = 6
No. of yellow marbles = 2

Total Marbles in Box = 15

Let S be the Sample Space
n(S) - No. of Ways of drawing 2 Marbles from the box of 15 marbles
n(S) = 15C2 = 15×7 = 105

Let E be the Event that the drawn 2 marbles are either blue or yellow.

Possible Cases for Event E occurrence:

1. Both Marbles are blue(E1)
P(E1) = 3C2/15C2 = (3×2)/(15×14) = 1/35

2. Both are yellow(E2)
P(E2)= 2C2/15C2 = 2/(15×14) = 1/105

3. One is Blue and Other is yellow(E3)
p(E3) = (3C1 × 2C1)/15C2= (2×3×2)/(15×14) = 2/35

E= E1 + E2 + E3
P(E) = n(E1) + n(E2) + n(E3)


We have,
P(E) = No. of favourable outcomes/No. of Total Outcomes

P(E) = (1/35) + (1/105) + 2/35
= (3/35) + (1/105)
= 1/35(3+(1/3))
= 10/(3×35)
= 2/21

•°• Required probability = 2/21

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