Question

One technique of killing cancer cells involves inserting microscopic synthetic rods, called carbon nanotubules, into the cell. When the rods are exposed to near-infrared light from a laser, they heat up, killing the cell, while cells without rods are left unscathed (Wong et al., 2005). Suppose that five nanotubules are inserted in a single cancer cell. Independently of each other, they become exposed to near-infrared light with probabilities 0.2, 0.4, 0.3, 0.6, and 0.5. What is the probability that the cell will be killed

Answer 1

Answer:

The probability that the cell will be killed is 0.9328.

Step-by-step explanation:

We are given that five nanotubules are inserted in a single cancer cell. Independently of each other, they become exposed to near-infrared light with probabilities 0.2, 0.4, 0.3, 0.6, and 0.5.

Let the event that a cell is killed be 'A' and the event where the ith nanotubule kill the cell be '$\text{B}_i$'.

This means that the cell will get killed if $\text{B}_1 \bigcup \text{B}_2 \bigcup \text{B}_3 \bigcup \text{B}_4 \bigcup \text{B}_5$ happens. This represents that the cell is killed if nanotubule 1 kills the cell, or nanotubule 2 kills the cell, and so on.

Here, P($\text{B}_1$) = 0.2, P($\text{B}_2$) = 0.4, P($\text{B}_3$) = 0.3, P($\text{B}_4$) = 0.6, P($\text{B}_5$) = 0.5.

So, the probability that the cell will be killed is given by;

P(A)= $1 - [(1 - P(\text{B}_1)) \times (1 - P(\text{B}_2)) \times (1 - P(\text{B}_3)) \times (1 - P(\text{B}_4)) \times (1 - P(\text{B}_5))]$

P(A) = $1 - [(1 - 0.2) \times (1 - 0.4) \times (1 - 0.3) \times (1 - 0.6) \times (1 - 0.5)]$

P(A) = $1 - (0.8 \times 0.6 \times 0.7 \times 0.4 \times 0.5)$

P(A) = 1 - 0.0672 = 0.9328

Hence, the probability that the cell will be killed is 0.9328.