MHT-CET Binomial Distribution — practice questions
30 free MCQs with worked solutions. Tap any question for the answer + explanation, or practice them all in the app.
Practice MHT-CET Binomial Distribution in the app →Three seeds are sown. Each independently germinates with probability 0.5. If $X$ = number that germinate, thenA **Bernoulli trial** is one in which:If $X im B(n, p)$, the probability of exactly $x$ successes is:For a Binomial distribution $X im B(n, p)$, the **mean** (expected value) is:For a Binomial distribution $X im B(n, p)$, the **variance** is:The **standard deviation** of $X im B(n, p)$ is:Drawing balls with **replacement** from an urn:If a fair coin is tossed $n$ times, the probability of any exact pattern (like all heads) is:$P(X \geq 1) = 1 - P(X = 0)$ is an example of the:If a die is thrown and 'getting an odd number' is success, then $p$ and $q$ are:A fair coin is tossed 10 times. The probability of getting **exactly 6 heads** is:A die is thrown 6 times. 'Getting an odd number' is a success. The probability of **exactly 5 successes** is:A pair of dice is thrown 4 times. 'Getting a doublet' is a success. The probability of **exactly 2 successes**Ten eggs drawn with replacement from a lot of 10% defective. Probability that **at least one** is defective:If $X im B(10, 0.4)$, then $E(X)$ and $\text{Var}(X)$ are respectively:If $E(X) = 6$ and $\text{Var}(X) = 4.2$ for $X im B(n, p)$, then $n$ and $p$ are:A bag has balls marked 0–9. Four balls drawn with replacement. Probability that **none** is marked '0' is:A multiple-choice exam has 5 questions, each with 3 options. Probability that a student gets at least 4 correcIf $X im B(5, p)$ and $P(X = 4) = P(X = 3)$, then $p$ equals:If $X im B(4, p)$ and $2P(X = 3) = 3P(X = 2)$, then $p$ equals:A die is thrown 100 times; 'even number' is success. The standard deviation of the number of successes is:Mean and variance of a binomial distribution are 4 and 2 respectively. The probability of **exactly 2** succesThe probability that a shooter hits a target is 3/4. The minimum number of shots required so that the probabilMean and variance of a binomial distribution are 18 and 12 respectively. The value of $n$ is:If $X im B(4, p)$ and $P(X = 0) = 16/81$, then $P(X = 4)$ equals:Probability of bomb hitting target = 0.8. Out of 10 bombs dropped, probability that **exactly 2 miss** equals If success probability in a single trial is 0.01, the minimum number of trials needed so that $P(\text{at leasIn a 5-trial binomial distribution, $P(X = 1) = 0.4096$ and $P(X = 2) = 0.2048$. The probability of success $pA bulb-factory has 5% defectives. In a random sample of 10 bulbs, the probability of **not more than one** defA multiple-choice exam has 10 questions each with 5 options. The probability that a student getting **8 or mor