Question

Five mangoes and 4 apples are in a box. If two fruits are chosen at random, find the probability that 

(i) one is a mango and the other is an apple 

(ii) both are of the same variety.

Answer 1

(i) Given that, 5 mangoes and 4 apples are in the box.

Two fruits out of 9 can be chosen in ⁹C₂ = 36 ways.

One mango and one apple can be chosen in ⁴C₁ × ⁵C₁ = 20 ways

∴ Probability = \(\frac{20}{36} = \frac 59\)

(ii) Let S be the sample space, A be the event of taking 2 mangoes and B be the event of taking 2 apples

∴ n(s) = 9C₂

= \(\frac {9\times 8}{ 1 \times 2}\)

= 9 × 4

= 36

n(A) = 5C₂

= \(\frac{5\times 4}{1\times 2}\)

= 10

n(B) = 4C₂

= \(\frac{4 \times 3}{1 \times 2}\)

= 6

P(taking 2 fruits are of the same colour)

= P(A or B)

= P(A ∪ B)

= P(A) + P(B)

\(= \frac{n(A)}{n(S)} + \frac{n(B)}{n(S)}\)

\(= \frac{10}{36} + \frac 6{36}\)

\(= \frac{10 + 6}{36}\)

\(= \frac{16 }{36}\)

\(= \frac 49\)

Answer 2

(i) Mangoes (M) = 5 

Apples (A) = 4 

Total = 5 + 4 = 9 

P(mango) = P(M) = 5/9 

P(A) = 4/9. 

When two Suits are chosen at random 

P(one mango and one Apple) = P(MA or AM) 

= P(M)P (A) × 2!

= (5/9) x (4/8) x 2! = 5/9 

(PCM) = 5/9 (Sell from 5 mangoes and set 1 from a fruit)

P(A) = 4/8 (Sell 1 from 4 apples and set 1 from the remaining 8 fruits)

(ii) P(MMorAA) 

= P(M) P(M) + P(A) P(A)