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A survey of people in a given region showed that 20% were smokers. The probability of death due to lung cancer, given that a person smoked, "was 10 times the probability of death due to lung cancer, given that a person did not smoke. If the probability of death due to lung cancer in the region is 0.006. What is the probability of death due to lung cancer given that a person is a smoker?

1/140
1/70
3/140
1/10

142.
Suppose a machine produces metal parts that contain some defective parts with probability 0.05. How many parts should be produced in order that the probability of atleast one part being defective is 1/2 or more? (Given that, log₁₀ 95 = 1.977 and log₁₀2 = 0.3)

11

12

15

14

Given probability of defective part = 0.005 = $\frac{1}{20}$

Probability of non-defective part = 1 - 0.005 = 0.95 = $\frac{19}{20}$

We know that, P (X = r) = ${}^{n}C_{r} p^{r} q^{n - r}$

where, $p = \frac{1}{20}, q = \frac{19}{20}, r \geq 1 and n = ?$

Also, $P (X \geq 1) \geq \frac{1}{2}$

$\Rightarrow 1 - P (X = 0) \geq \frac{1}{2} \Rightarrow 1 - {}^{n}C_{0} {(\frac{1}{20})}^{0} {(\frac{19}{20})}^{n - 0} \geq \frac{1}{2} \Rightarrow 1 - \frac{1}{2} \geq {(\frac{19}{20})}^{n} \Rightarrow \frac{1}{2} \geq {(\frac{19}{20})}^{n} \Rightarrow \frac{1}{2} \geq {(\frac{95}{100})}^{n} \Rightarrow \log {(2)}^{- 1} \geq n [\log (95) - \log (100)] \Rightarrow - \log (2) \geq n [\log (95) - 2] \Rightarrow - (0.3) \geq n [1.977 - 2] \Rightarrow n \geq \frac{0.3}{0.023} \Rightarrow n \geq \frac{300}{23} \Rightarrow n \geq 13.04 ∴ n = 14, 15$

143.

There is a group of 265 persons who like either singing or dancing or painting. In this group 200 like singing, 110 like dancing and 55 like painting. If 60 persons like both singing and dancing, 30 like both singing and painting and 10 like all three activities, then the number of persons who like only dancing and painting is

144.

For two events A and B, let P(A) = 0.7 and P(B) = 0.6. The necessarily false statement(s) is/are

$P (A \cap B) = 0.35$
$P (A \cap B) = 0.45$
$P (A \cap B) = 0.65$
$P (A \cap B) = 0.28$

145.

An objective type test paper has 5 questions. Out of these 5 questions, 3 questions have four options each (a, b, c, d) with one option being the correct answer. The other 2 questions have two options each, namely true and false. A candidate randomly ticks the options. Then, the probability that he/she will tick the correct option in atleast four questions, is

$\frac{5}{32}$
$\frac{3}{128}$
$\frac{3}{256}$
$\frac{3}{64}$

146.

Two coins are available, one fair and the other two headed. Choose a coin and toss it once; assume that the unbiased coin is chosen with probability $\frac{3}{4}$ . Given that the outcome is head, 4 the probability that the two headed coin was chosen, is

$\frac{3}{5}$
$\frac{2}{5}$
$\frac{1}{5}$
$\frac{2}{7}$

147.

A person draws out two balls successively from a bag containing 6 red and 4 white balls. The probability that at least one of them will be red, is

$\frac{78}{90}$
$\frac{30}{90}$
$\frac{48}{90}$
$\frac{12}{90}$

Short Answer Type

148.

If an unbiased coin is tossed n times. Find the probability that head appears an odd number of times.

Multiple Choice Questions

149.

The mean and the variance of a binomial distribution are 4 and 2, respectively. Then, the probability of 2 succeses is

28/256
$\frac{219}{256}$
128/256
37/256

150.

An unbiased coin is tossed n times. Let X denotes the number of times head occurs. If P(X = 4 ), P(X = 5) and PX = 6) are in AP, then the value of n can be