Một hộp đựng 10 bi, trong đó có 4 bi xanh và 6 bi đỏ. Lấy ra 1 bi và không hoàn lại, tiếp tục lấy ra 1 bi nữa. Gọi : “Bi lấy được lần 1 màu xanh”, : “Bi lấy được lần 2 màu đỏ” tính P(AB)

Question

...T-h-i-ê-n..L-ý..C-ó..B-ồ...|.|hg2 · Accepted Answer

Do lấy hai lần khác nhau nên biến cố A và B độc lập
$\displaystyle \begin{array}{{>{\displaystyle}l}}
P( A) =\frac{C_{4}^{1}}{C_{10}^{1}} =\frac{2}{5}\
P( B) =\frac{C_{6}^{1}}{C_{9}^{1}} =\frac{2}{3}\
\Longrightarrow P( AB) =P( A) .P( B) =\frac{2}{5} .\frac{2}{3} =\frac{4}{15}
\end{array}$

Timi · Answer

To solve this problem, we can use the concept of conditional probability. The probability of event A (getting a blue ball on the first draw) and then event B (getting a red ball on the second draw) can be calculated using the formula:

$$ P(AB) = P(A) \times P(B|A) $$

Where:
- $ P(A) $ is the probability of getting a blue ball on the first draw
- $ P(B|A) $ is the conditional probability of getting a red ball on the second draw given that a blue ball was drawn on the first draw

First, let&#x27;s calculate $ P(A) $, the probability of getting a blue ball on the first draw:
$$ P(A) = \frac{\text{Number of blue balls}}{\text{Total number of balls}} = \frac{4}{10} = 0.4 $$

Next, we need to calculate $ P(B|A) $, the conditional probability of getting a red ball on the second draw given that a blue ball was drawn on the first draw. Since we do not replace the first ball, after drawing one blue ball, there are 9 balls left in total, with 6 red balls remaining.

Therefore,
$$ P(B|A) = \frac{\text{Number of red balls left}}{\text{Total number of balls left}} = \frac{6}{9} = 0.6666666666666666$$

Now we can find $P(AB)$:
$$P(AB)=P(A)\times P(B|A)=0.4\times0.6666666666666666=0.26666666666666666$$

So, 
$$P(AB)=0.26666666666666666$$

Hence, the final answer for $P(AB)$, which represents the probability of getting a blue ball on the first draw and a red ball on the second draw, is 0.2667 or approximately 26.67%.