Question

There are two boxes a and



b. box a has 5 oranges and 6 apples in it and box b contains 3 apples and 4 oranges in it. a fruit is taken from a and placed in b, after which a fruit is then transferred from b to



a. what is the probability that the configuration of boxes does not change due to the transfers

Answer 1

Answer:

49/88

Step-by-step explanation:

As there can be two cases either u take orange from first box 5/11 and again take orange from second box 5/8 or u can take one apple from first box 6/11 and second apple from second box 4/8

so if we add these two

5/11*5/8+6/11*4/8 will get the answer 49/88

Answer 2

Answer:

The required probability is $\frac{49}{88}$ .

Step-by-step explanation:

Case I, When one orange fruits is transferred from A to B and B to A.

$P(E)_{1}$ = ( ⁵C₁ ÷ ¹¹C₁ ) × ( ⁵C₁ ÷ ⁸C₁ )

          = $\frac{25}{88}$

Case II, When one apple is transferred from A to B, and B to A,

$P(E)_{2}$ = ( ⁶C₁ ÷ ¹¹C₁ ) × ( ⁴C₁ ÷ ⁸C₁ )

          = $\frac{24}{88}$

∴ Required probability = $P(E)_{1}$ + $P(E)_{2}$

          = $\frac{25}{88}$ + $\frac{24}{88}$

          = $\frac{49}{88}$ (Ans.)