Question

Suppose researchers determined that a new drug has a 55% chance of preventing a certain flu strain if the drug is administered to 10 male subjects, what is the probability that the drug it will be effective in preventing the flu strain for fewer than three of the male subjects? Select one: a. 0.09956 b. 0.101995 OC 0.027392 d. None of these

Answer 1

1) option  b

Explanation:- Let X denotes number of males subject in which drugs are administered for preventing a certain flu strain and X is binomial variate

Then we have to find probability that the drug will be effective in preventing the flu strain for fewer than 3 of male subjects

i .e P(X$\leqslant$3).

Given ,n=10,p=0.55 ; 0$\leq$X$\leq$10

Then

P(X$\leqslant$3)= nCx$p^{x}$(1-$p^{x}$)

=P(X=0)+P(X=1)+P(X=2)+P(X=3)

=0.0003+0.0042+0.0229+0.0746

=0.10995

3)Option  b

If two events are Independent then the probability that both the events occurs is the product of their individual probabilities

If A and B are two independent events then

I. e P(A and B)= P(A) * P(B)

4)Option a

Probability value lies between 0 and 1.

Therefore largest possible probability value is 1

2) option b

Out of 6 members of team , n=6

r=3,no. Of items in each permutation

We have to select a team leader, team secretary, team president

a permutation is an ordered combination.

The number of permutations of n objects taken r at a time is determined by the following formula:

P(n,r)=n!/(n−r)!

= 6!/3!

=120